Class 11

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 10 Straight Lines Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 10 Straight Lines Miscellaneous Exercise

Question 1.
Find the values of k for which the line (k – 3) x – (4 – k²)y + k² – 7k + 6 = 0 is
(a) Parallel to the x-axis
(b) Parallel to the y-axis
(c) Passing through the origin.
Answer.
The given equation of line is
(k – 3) x – (4 – k²)y + k² – 7k + 6 = 0
(a) If the given line is parallel to the x-axis, then
Slope of the given line = Slope of the x-axis The given line can be written as
(4 – k²)y = (k – 3)x + k² – 7k + 6 = 0
y =\(\frac{(k-3)}{\left(4-k^{2}\right)} x+\frac{k^{2}-7 k+6}{\left(4-k^{2}\right)}\), which is of the form y = mx + c.

Slope of the given line = \(\frac{(k-3)}{\left(4-k^{2}\right)}\)

Slope of the x-axis = 0
\(\frac{(k-3)}{\left(4-k^{2}\right)}\) = 0

⇒ k – 3 = 0
⇒ k = 3
Thus, if the given line is parallel to the x-axis, then the value of k is 3.

(b) If the given line is parallel to the y-axis, it is vertical. Hence, its slope will be undefined.
The slope of the given line is \(\frac{(k-3)}{\left(4-k^{2}\right)}\)

Now, \(\frac{(k-3)}{\left(4-k^{2}\right)}\) is undefined at k² = 4

k² = 4
⇒ k = ± 2
Thus, if the given line is parallel to the y-axis, then the value of k is ± 2.

(c) If the given line is passing through the origin, then point (0, 0) satisfies the given equation of line.
(k – 3) (0) – (4 – k²) (0) + k² – 7k + 6 = 0
k² – 7k + 6 = 0
k² – 6k – k + 6 = 0
(k – 6) (k – 1) = 0 k = 1 or 6
Thus, if the given line is passing through the origin, then the value of k is either 1 or 6.

Question 2.
Find the value of 6 and p, if the equation x cos θ + y sin θ = p is the normal form of the line √3x + y + 2 = 0.
Answer.
The equation of the given line is √3x + y + 2 = 0.
This equation can be reduced as √3x + y + 2 = 0
⇒ On dividing both sides by \(\frac{-\sqrt{3} x-y=2}{(-\sqrt{3})^{2}+(-1)^{2}}\) = 2, we obtain

\(-\frac{\sqrt{3}}{2} x-\frac{1}{2} y=\frac{2}{2}\)

⇒ \(\left(-\frac{\sqrt{3}}{2}\right) x+\left(-\frac{1}{2}\right) y\) = 1

On comparing equation (i) to x cos θ + y sin θ = p, we obtain
cos θ = – \(\frac{\sqrt{3}}{2}\), sin θ = – \(\frac{1}{2}\) and p = 1
Since the values of sin θ and cos θ are negative, θ = π + \(\frac{\pi}{6}\) = \(\frac{7 \pi}{6}\).
Thus, the respective values of θ and p are \(\frac{7 \pi}{6}\) and 1.

Question 3.
Find the equation of the lines, which cut-off intercepts on the axes whose sum and product are 1 and -6, respectively.
Answer.
Let the intercepts form of line be \(\frac{x}{a}+\frac{y}{b}\) = 1,
then a + b = 1 and ab = – 6.
b = 1 – a and a (1 – a) = – 6
⇒ a² – a – 6 = 0
⇒ (a – 3) (a + 2) = 0
∴ a = 3, – 2

Case I:
If a = 3, then b = – \(\frac{6}{a}\)
= – \(\frac{6}{3}\) = – 2
∴ Equation of line is \(\frac{x}{3}+\frac{y}{-2}\) = 1
⇒ 2x – 3y – 6 = 0.

Case II:
If a = – 2, then b = \(\frac{-6}{-2}\) = 3
∴ Equation of line is \(\frac{x}{-2}+\frac{y}{3}\) = 1
⇒ 3x – 2y + 6 = 0.

Question 4.
What are the points on the y-axis whose distance from the line \(\frac{x}{3}+\frac{y}{4}\) = 1 is 4 units.
Answer.
Let (0, b) be the point on they-axis whose distance from line \(\frac{x}{3}+\frac{y}{4}\) = 1 is 4 units.
The given line can be written as 4x + 3y – 12 = 0 ……………..(i)
On comparing equation (i) to the general equation of line Ax + By + C = 0,we obtain A = 4, B = 3 and C = – 12.
It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by
d = \(\)
Therefore, if (0, b) is the point on the y-axis whose distance from line \(\frac{x}{3}+\frac{y}{4}\) = 1 is 4 units, then:
4 = \(\frac{|4(0)+3(b)-12|}{\sqrt{4^{2}+3^{2}}}\)

4 = \(\frac{|3 b-12|}{5}\)
⇒ 20 = |3b – 12|
20 = ±(3b – 12)
20 = (3b – 12) or 20 = – (3b – 12)
3b = 20 + 12 or 3b = – 20 + 12
b = \(\frac{32}{3}\) or b = \(\frac{8}{3}\).
Thus, the required points are (o, \(\frac{32}{3}\)) and (o, \(\frac{8}{3}\)).

Question 5.
Find perpendicular distance from the origin of the line joining the points (cos θ, sin θ) and (cos Φ, sin Φ).
Answer.
Equation of the line joining the points (cos θ, sin θ) are (cos Φ, sin Φ) is given by

Question 6.
Find the equation of the line parallel to y-axis and drawn through the point of intersection of the lines x – 7y + 5 = 0 and 3x + y = 0.
Answer.
The point of intersection (x₁, y₁) of the lines x – 7y + 5 = 0 and 3x + y = 0 is obtained by solving these equations.
Putting y = – 3x in x – 7y + 5 = 0
⇒ x – 7 (- 3x) + 5 = 0,
⇒ x + 21x + 5 = 0
⇒ 22x + 5 = 0
⇒ x = \(\frac{5}{22}\)
Also, y = \(\frac{15}{12}\)
⇒ (x₁, y₁) = \(\left(\frac{-5}{22}, \frac{15}{12}\right)\)
Let a line parallel to y-axis through the point (x₁, y₁) is x = x₁
Here x₁ = – \(\frac{5}{22}\)
∴ The equation of the line parallel to y-axis passing through the point of intersection (x₁, y₁) of given lines is x = – \(\frac{5}{22}\) or 22x + 5 = 0.

Question 7.
Find the equation of a line drawn perpendicular to the line \(\frac{x}{4}+\frac{y}{6}\) = 1 through the point, where It meets the y-axis.
Answer.
Given equation of line is \(\frac{x}{4}+\frac{y}{6}\) = 1

\(\frac{3 x+12}{12}\) = 1

3x + 2y = 12 ……………..(i)
If line (i) meet the Y-axis, then put x = 0 in eq. (i), we get
0 + 2y = 12
⇒ y = 6
∴ Point is (0, 6).
Slope of line (i) is, m₁ = – \(\frac{3}{2}\)
Slope of line perpendicular to line (i) is,
m₂ = – \(-\frac{1}{m_{1}}=\frac{-1}{(-3 / 2)}=\frac{2}{3}\)

Now, equation of line having slope and passing through (0, 6) is given by
y – y₁ = m (x – x₁)
⇒ y – 6 = \(\frac{2}{3}\) (x – 0)
⇒ 3y – 18 = 2x
⇒ 2x – 3y + 18 = 0
Which is required equation of line.

Question 8.
Find the area of the triangle formed by the lines y – x = 0, x + y = 0 and x – k= 0.
Answer.

Let ABC be the triangle, whose sides are
AB: x – k = 0 ……………..(i)
BC: y – x = 0 ………………(ii)
and AC: x + y = 0 ………………(iii)
On solving eqs. (i) and (iii), we get Coordinates of A i.e.,(k, – k)
On solving eqs. (i) and (ii), we get Coordinates of B i.e.,(k, k)
On solving eqs. (ii) and (iii), we get Coordinates of C i.e. (0, 0)
∴ Area of ∆ABC = \(\frac{1}{2}\) {x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)
= \(\frac{1}{2}\) {k (k – 0) + k (0 + k) + 0 (- k – k)}
[∵ (x₁, y₁) = (k,- k), (x₂, y₂) = (k, k) and (x₃, y₃) = (0, 0)]
= \(\frac{1}{2}\) {k² + k²} = \(\frac{1}{2}\) k²
= k²

Question 9.
Find the value of p so that the three lines 3x + y – 2 = 0 px + 2y – 3 = 0 and 2x – y – 3 = 0 may intersect at one point.
Answer.
The equations of the given lines are
3x + y – 2 = 0 …………(i)
px + 2y – 3 = 0 …………..(ii)
2x – y – 3 = 0 ……………..(iii)
On solving equations (i) and (iii), we obtain x = 1 and y = – 1.
Since these three lines may intersect at one point, the point of intersection of lines (i) and (iii) will also satisfy line (ii). p(1) + 2(- 1) – 3 = 0
p – 2 – 3 = 0
⇒ p = 5
Thus, the required value of p is 5.

Question 10.
If three lines whose equations are y = m₁x + c₁, y = m₂x + c₂ and y = m₃x + c₃ are concurrent, then show that m₁ (c₂ – c₃) + m₂ (c₃ – c₁) + m₃ (c₁ – c₂) = 0.
Answer.
The equations of the given lines are
y = m₁x + c₁ ………………(i)
y = m₂x + c₂ ……………..(ii)
y = m₃x + c₃ ………………(iii)
On subtracting equation (i) from (ii), we obtain 0 = (m₂ – m₁) x + (c₂ – c₁)
⇒ (m₁ – m₂)x = c₂ – c₁
⇒ x = \(\frac{c_{2}-c_{1}}{m_{1}-m_{2}}\)

On substituting this value of x in eq. (i), we obtain

y = \(m_{1}\left(\frac{c_{2}-c_{1}}{m_{1}-m_{2}}\right)+c_{1}=\frac{m_{1} c_{2}-m_{1} c_{1}}{m_{1}-m_{2}}+c_{1}\)

= \(\frac{m_{1} c_{2}-m_{1} c_{1}+m_{1} c_{1}-m_{2} c_{1}}{m_{1}-m_{2}}=\frac{m_{1} c_{2}-m_{2} c_{1}}{m_{1}-m_{2}}\)

∴ \(\left(\frac{c_{2}-c_{1}}{m_{1}-m_{2}}, \frac{m_{1} c_{2}-m_{2} c_{1}}{m_{1}-m_{2}}\right)\) is the point of intersection of lines (i) and (ii).

It is given that lines (i), (ii) and (iii) are concurrent.
Hence the point of intersection of lines (i) and (ii) will also satisfy equation (iii).

\(\frac{m_{1} c_{2}-m_{2} c_{1}}{m_{1}-m_{2}}=m_{3}\left(\frac{c_{2}-c_{1}}{m_{1}-m_{2}}\right)+c_{3}\) \(\frac{m_{1} c_{2}-m_{2} c_{1}}{m_{1}-m_{2}}=\frac{m_{3} c_{2}-m_{3} c_{1}+c_{3} m_{1}-c_{3} m_{2}}{m_{1}-m_{2}}\)

m₁c₂ – m₂c₁ – m₃c₁ – c₃m₁ + c₃m₂ = 0

m₁ (c₂ – c₃) + m₂ (c₃ – c₁) + m₃ (c₁ – c₂) = 0.
Hence proved.

Question 11.
Find the equation of the lines through the point (3, 2) which make an angle of 45° with the line x – 2y = 3.
Answer.
Let the slope of the required line be m₁.
The given line can be written as y = \(\frac{1}{2} x-\frac{3}{2}\) which is of the form y = mx + c.
Slope of the given line = m₂ = \(\frac{1}{2}\)
It is given that the angle between the required line and line x – 2y = 3 is 45°
We know that if θ is the acute angle between lines l₁ and l₂ with slopes m₁ and m₂ respectively, then

Case I : m₁ = 3
The equation of the line passing through (3, 2) and having a slope of 3 is:
y – 2 = 3 (x – 3)
y – 2 = 3x – 9
3x – y = 7

Case II: m₁ = – \(\frac{1}{3}\).
The equation of the line passing through (3, 2) and having a slope of – \(\frac{1}{3}\) is
y – 2 = – \(\frac{1}{3}\) (x – 3)
3y – 6 = – x + 3
x + 3y = 9.
Thus, the equations of the lines are 3x – y = 7 and x + 3y = 9.

Question 12.
Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes.
Answer.

The given lines are
2x – 3y = – 1 …………….(i)
4x + 7y = 3 ……………(ii)
Multiplying eq. (i) by 2
4x – 6y = – 2 ………….(iii)
Subtracting eq. (iii) from eq. (ii), we get
13y = 5
⇒ y = \(\frac{13}{5}\)
Putting the value of y in eq. (i),
2x – \(\frac{3 \times 5}{13}\) = – 1
2x = – 1 + \(\frac{15}{13}=\frac{2}{13}\)
x = \(\frac{1}{13}\)
∴ Given lines intersect at P(\(\frac{1}{13}\), \(\frac{5}{13}\))
PA and PB are the lines that make equal intercepts on the axes.
They make angles of 135° with positive direction of x-axis.
Their slopes are tan 135° and tan 45° i.e., – 1 and 1 respectively.
∴ Equation of PA is y – \(\frac{5}{13}\) = (- 1) × (x – \(\frac{1}{13}\))
or 13y – 5 = – 13x + 1
13x + 13y – 6 = 0
Similarly, equation of PB is y – \(\frac{5}{13}\) = 1 × (x – \(\frac{5}{13}\))
⇒ 13y – 5 = 13x – 1
13x – 3y + 4 = 0.

Question 13.
Show that the equation of the line passing through the origin and nisildng an angle θ with the line y = mx + c is
\(\frac{y}{x}=\frac{m \pm \tan \theta}{1 \mp m \tan \theta}\)
Answer.
Slope of line y = mx + c is m.
Let M be the slope of required line, then
tan θ = \(\left|\frac{M-m}{1+m M}\right|=\pm\left(\frac{M-m}{1+M m}\right)\)

Case I:
Taking ‘+‘ sign,
tan θ = \(\frac{M-m}{1+m M}\)
Then, tan θ + Mm . tan θ = M – m

Question 14.
In what ratio, the line joining (- 1, 1) and (5, 7) is divided by the line x + y = 4?
Answer.
The equation of the line joining the points (- 1, 1) and (5, 7) is given by
y – 1 = \(\frac{7-1}{5+1}\) (x + 1)
= \(\frac{6}{6}\) (x + 1)
x – y + 2 = 0 ………….(i)
The equation of the given line is x + y – 4 = 0 ……………..(ii)
The point of intersection of lines (i) and (ii) is given by x = 1 and y = 3.
Let point (1, 3) divide the line segment joining (- 1, 1) and (5, 7) in the ratio 1 : k. Accordingly, by section formula.
(1, 3) = \(\left(\frac{k(-1)+1(5)}{1+k}, \frac{k(1)+1(7)}{1+k}\right)\)

(1, 3) = \(\left(\frac{-k+5}{1+k}, \frac{k+7}{1+k}\right)\)

\(\frac{-k+5}{1+k}\) = 1

∴ – k + 5 = 1 + k
⇒ 2k = 4
⇒ k = 2.
Thus, the line joining the points (- 1, 1) and (5, 7) is divided by line x + y = 4 in the ratio 1 : 2.

Question 15.
Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line 2x – y = 0.
Answer.
The given lines are
2x – y = 0 ………………..(i)
4x + 7y + 5 = 0 ………………..(ii)
A(1, 2) is a point on line (i).
Let B be the point of intersection of lines (i) and (ii).
On solving equations (i) and (ii), we obtain x = – \(\frac{5}{18}\) and y = – \(\frac{5}{9}\).
∴ Coordnates of point B = (- \(\frac{5}{18}\), – \(\frac{5}{9}\))

Thus, the required distance is \(\frac{23 \sqrt{5}}{18}\) units.

Question 16.
Find the direction in which a straight line must be drawn through the point (- 1, 2) so that its point of intersection with the line x + y = 4 may be at a distance of 3 units from this point.
Answer.
Any line passing through P(- 1, 2) is
y – 2 = m(x + 1)
where m is its slope
y = mx + m + 2 ……………(i)
Putting the value of ‘y’ in x + y = 4, we get
x + mx +m + 2 = 4
⇒ (1 + m) x = 4 – 2 – m
⇒ x = \(\frac{2-m}{1+m}\)
From eq.(i),
y = m (\(\frac{2-m}{1+m}\)) + m + 2
Now from the diagram, it is clear that the point of intersection of the other two legs of the right triangle having AB as the hypotenuse can be either P or Q.

Case I: When ∠APB is taken,
The perpendicular sides in ∠APB are AP and PB.
Now, side PB is parallel to x-axis and at a distance of 1 unit above x-axis.
So, equation of PB is y = 1 or y – 1 = 0.
The side AP is parallel to y-axis and at a distance of 1 unit on the right of y-axis.
So, equation of AP is x = 1 or x – 1 = 0.

Case II: When ∠AQB is taken.
The perpendicular sides in ∠AQB are AQ and QB.
Now, side AQ is parallel to x-axis and at a distance of 3 units above x – axis.
So, equation of AQ is y = 3 or y – 3 = 0.
The side QB is parallel to y-axis and at a distance of 4 units on the left of y-axis.
So, equation of QB is x = – 4 or x + 4 = 0.
Hence, the equation of the legs are: x = 1, y = 1 or x = – 4, y = 3.

Question 18.
Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.
Answer.
Let AB be the line x + 3y = 7 and the image P(3, 8) of P(3, 8) be Q(x₁, y₁)middle point at PQ

M \(\left(\frac{x_{1}+3}{2}, \frac{y_{1}+8}{2}\right)\) lies on AB.

∴ \(\left(\frac{x_{1}+3}{2}\right)+3\left(\frac{y_{1}+8}{2}\right)\) = 7
x₁ + 3 + 3y₁ + 24 = 14
⇒ x₁ + 3y₁ + 13 = 0 ……………….(i)
slope of AB = – \(\frac{1}{3}\),
Slope of PQ = \(\frac{y_{1}-8}{x_{1}-3}\)
AB ⊥ PQ

\(\left(-\frac{1}{3}\right)\left(\frac{y_{1}-8}{x_{1}-3}\right)\) = – 1

⇒ y₁ – 8 = 3 (x₁ – 3) = 3x₁ – 9
⇒ y₁ = 3x₁ – 1 ………….(ii)
Putting the value of y₁ in eq. (1), we get
x₁ + 3 (3x₁ – 1) + 13 = 0
10x₁ + 10 = 0
x₁ = – 1
Putting the value of x₁ in (ii), we get
y₁ = – 3 – 1 = – 4
∴ The point Q, the image of P is (- 1, 4).

Question 19.
If the lines y = 3x + 1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.
Answer.
The equations of the given lines are
y = 3x + 1
2y = x + 3 …………..(ii)
y = mx + 4 …………….(iii)
Slope of line (i), m₁ = 3.
Slope of line (ii), m₂ = \(\frac{1}{2}\)
Slope of line (iii), m₃ = m.
It is given that lines (i) and (ii) are equally inclined to line (iii).
This means that the angle between lines (i) and (iii) equals the angles between lines (ii) and (iii).

Question 20.
If sum of the perpendicular distances of a variable point P(x, y) from the lines x + y – 5 = 0 and 3x – 2y + 7 = 0 is always 10. Show that P must move on a line.
Answer.
The equations of the given lines are
x + y – 5 = 0 …………….(i)
3x – 2y + 7 = 0 ………………(ii)
The perpendicular distances of P(x, y) from lines (j) and (ii) are respectively given by
d₁ = \(\frac{|x+y-5|}{\sqrt{(1)^{2}+(1)^{2}}}\) and

d₂ = \(\frac{|3 x-2 y+7|}{\sqrt{(3)^{2}+(-2)^{2}}}\)

i.e., d₁ = \(\frac{|x+y-5|}{\sqrt{2}}\)

d₂ = \(\frac{|3 x-2 y+7|}{\sqrt{13}} .\)

It is given that d₁ + d₂ = 10

∴ \(\frac{|x+y-5|}{\sqrt{2}}+\frac{|3 x-2 y+7|}{\sqrt{13}}\) = 10

⇒ √13 |x + y – 5| + √2 |3x – 2y + 7| – 10√26 = 0
⇒ √13 (x + y – 5) + √2 (3x – 2y + 7) – 10√26 = 0
[Assuming (x + y – 5) and (3x – 2y + 7) are positive]
⇒ √13x + √13y – 5√13 + 3√2x – 2√2y + 7√2 – 10√26 = 0
⇒ x (√13 + 3√2) + y (√13 – 2√2) + (7√2 – 5√13 – 10√26) = 0
which is the equation of a line.
Similarly, we can obtain the equation of line for any signs of (x + y – 5) and (3x – 2y + 7).
Thus, point P must move on a line.

Question 21.
Find equation of the line which is equidistant from parallel lines 9x+6y – 7 = 0 and 3x + 2y + 6 = 0.
Answer.
The given parallel lines are
9x + 6y – 7 = 0 …………….(i)
and 3x + 2y + 6 = 0 …………….(ii)
Multiplying (ii) by 3, we get
9x + 6y + 18 = 0 ………………….(iii)
Let the equations of (ii) line parallel to the lines (i) and (iii) is
9x + 6y + c = 0 …………..(iv)
Distance between (i) and (iv)
= \(\frac{|-7-c|}{\sqrt{9^{2}+6^{2}}}=\frac{|7+c|}{\sqrt{117}}\)
Distance between (iii) and (iv)
= \(\frac{|18-c|}{\sqrt{9^{2}+6^{2}}}\)
The third line being equidistant from the given two lines.
\(\frac{|7+c|}{\sqrt{117}}=\frac{|c+18|}{\sqrt{117}}\) or 2c = 11 or c = \(\frac{11}{2}\)
Putting this values of c in eq. (iv), we get
9x + 6y + \(\frac{11}{2}\) = 0
or 18x + 12y + 11 = 0
which is the equation of required line.

Question 22.
A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.
Answer.
In the figure, PA is the incident ray and AR is the reflected ray, which makes an angle 0 from the X-axis.

It is clear from the figure that AS ⊥ OX
It means AS bisect the ∠PAR.
Then, ∠PAS = ∠RAS
⇒ ∠RAX = ∠PAO = θ (let)
⇒ ∠XAP = 180° – θ
Slope of AR = tan θ
= \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

= \(\frac{3-0}{5-k}\) …………….(i)
[where, point A is (k, 0)]
Slope of AP = tan (180 – θ)
= – tan θ
= \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{2-0}{1-k}\) ……………(ii)
From eqs. (i) and (ii), we get
\(\frac{3}{5-k}=-\frac{2}{1-k}\)

⇒ 3 – 3k = – 10 + 2k
⇒ 5k = 13
⇒ k = \(\frac{13}{5}\)
Hence, the coordinates of A are (\(\frac{13}{5}\). 0).

Question 23.
Prove that the product of the lengths of the perpendiculars drawn from the points (\(\sqrt{a^{2}-b^{2}}\), 0) and (- \(\sqrt{a^{2}-b^{2}}\), 0) to the line \(\frac{x}{a}\) cos θ + \(\frac{y}{4}\) sin θ = 1 is b².
Answer.
The equation of the given line is \(\frac{x}{a}\) cos θ + \(\frac{y}{4}\) sin θ = 1
or bx cos θ + ay sin θ – ab = 0 ……………….(i)
Length of the perpendicular from point (\(\sqrt{a^{2}-b^{2}}\), 0) to the line (i) is
P₁ = \(\frac{\mid b \cos \theta\left(\sqrt{\left.a^{2}-b^{2}\right)}+a \sin \theta(0)-a b \mid\right.}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}\)

= \(\frac{\left|b \cos \theta \sqrt{a^{2}-b^{2}}-a b\right|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}\) ……………….(ii)
Length of the perpendicular from point (- \(\sqrt{a^{2}-b^{2}}\), 0) to line (ii) is
P₁ = \(\frac{b \cos \theta\left(-\sqrt{a^{2}-b^{2}}\right)+a \sin \theta(0)-a b}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}\)

= \(\frac{\left|b \cos \theta \sqrt{a^{2}-b^{2}}+a b\right|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}\)

On multiplying equations (ii) and (iii), we obtain

Question 24.
A person standfing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = O and
3x + 4y – 5 = 0 wants to reach the path whose equation Is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.
Answer.
The equations of the given lines are
2x – 3y + 4 = 0
3x + 4y – 5 = 0 …………..(ii)
6x – 7y + 8 = 0 …………(iii)
The person is standing at the junction of the paths represented by lines (i) and (ii).
On solving equations (i) and (ii), we obtain x = – \(\frac{1}{17}\) and y = \(\frac{22}{17}\)
Thus, the person is standing at point (- \(\frac{1}{17}\), \(\frac{22}{17}\))
The person can reach path (iii) in the least time if he walks along the perpendicular line to (iii) from point (- \(\frac{1}{17}\), \(\frac{22}{17}\)).
Slope of the line (iii) = \(\frac{6}{7}\).
∴ Slope of the line perpendicular to line (iii) = \(-\frac{1}{\left(\frac{6}{7}\right)}=-\frac{7}{6}\)
The equation of the line passing through and having a slope (- \(\frac{1}{17}\), \(\frac{22}{17}\)) and having a slope of – \(\frac{7}{6}\) is given by
\(\left(y-\frac{22}{17}\right)=-\frac{7}{6}\left(x+\frac{1}{17}\right)\)
6 (17y – 22) = – 7 (17x + 1)
102y – 132 = – 119x – 7
119x + 102y = 125
Hence, the path that the person should follow is 119x + 102y = 125.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 10 Straight Lines Ex 10.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3

Question 1.
Reduce the following equations into slope-intercept form and find their slopes and the y-intercepts.
(i) x + 7y = 0
(ii) 6x + 3y – 5 = 0
(iii) y = 0
Answer.
(i) The given equation is x + 7y = 0
It can be written as
y = – \(\frac{1}{7}\) x + 0 …………….(i)
This equation is of the form y = mx + c, where m = – \(\frac{1}{7}\) and c = 0.
Therefore, equation (i) is in the slope-intercept form, where the slope and the y-intercept are – \(\frac{1}{7}\) and 0 respectively.

(ii) The given equation is 6x + 3y – 5 = 0.
It can be wrirtenas y = \(\frac{1}{3}\) (- 6x + 5)
y = – 2x + \(\frac{5}{3}\)
This equation is of the form y = mx + c, where m = – 2 and c = \(\frac{5}{3}\)
Therefore, equation (ii) is in the slope-intercept form, where the slope and the y-intercept are – 2 and \(\frac{1}{3}\) respectively.

(iii) The given equation is y = 0.
It can be written as y = 0.x + 0 …………..(iii)
This equation is of the form y = mx + c, where m = 0 and c = 0.
Therefore, equation (iii) is in the slope-intercept form, where the slope and the y-intercept are 0 and 0 respectively.

Question 2.
Reduce the following equations into intercept form and find their intercepts on the axes.
(i) 3x + 2y – 12 = 0
(ii) 4x – 3y = 6
(iii) 3y + 2 = 0
Answer.
(i) The given equation is 3x + 2y -12 = 0
It can be written as 3x + 2y = 12
\(\frac{3 x}{12}+\frac{2 y}{12}\) = 1

i.e., \(\frac{x}{4}+\frac{y}{6}\) = 1

This equation is of the form \(\frac{x}{a}+\frac{y}{b}\) = 1, where a = 4 and b = 6.
Therefore, equation (i) is in the intercept form, where the intercepts on the x and y axes are 4 and 6 respectively.

(ii) The given equation is 4x – 3y = 6
It can be written as \(\frac{4 x}{6}-\frac{3 y}{6}\) = 1
\(\frac{2 x}{3}-\frac{y}{2}\) = 1

\(\frac{x}{\left(\frac{3}{2}\right)}+\frac{y}{(-2)}\) = 1
This equation is of the form \(\frac{x}{a}+\frac{y}{b}\) = 1, where a = \(\frac{3}{2}\) and b = – 2.
Therefore, equation (ii) is in the intercept form, where the intercepts on the x and y – axes are \(\frac{3}{2}\) and – 2 respectively.

(iii) The given equation is 3y +2 = 0.
It can be written as 3y = – 2
i.e., \(\frac{y}{\left(-\frac{2}{3}\right)}\) = 1 ……………….(iii)
This equation is of the form \(\frac{x}{a}+\frac{y}{b}\) = 1,where a = 0 and b = – \(\frac{2}{3}\).
Therefore, equation (iii) is in the intercept form, where the intercept on the y-axis is – \(\frac{2}{3}\) and it has no intercept on the x-axis.

Question 3.
Reduce the following equations into the nornml form. Find their perpendicular distance from the origin and nngle between perpendicular and positive direction of x-axis.
(i) x – √3y + 8 = 0
(ii) y – 2 = 0
(iii) x – y = 4
Answer.
(i) x – √3y + 8 = 0 or
x – √3y = – 8 or
– x + √3y = 8
Put r cos ω = – 1, r sin ω = √3
Squaring and adding r² = 1 + 3 = 4
∴ r = 2
cos ω = – \(\frac{1}{2}\), sin ω = \(\frac{\sqrt{3}}{2}\) is
∴ ω = 120 = \(\frac{2 \pi}{3}\)
∴ x cos w + y sin w = \(\frac{8}{2}\) = 4
∴ p = 4 and ω = \(\frac{2 \pi}{3}\)

(ii) y – 2 = 0
Comparing with Ax + By = C
A = 0, B = 1, r = \(\sqrt{0+1^{2}}\) = 1
cos ω = 0, sin ω = 1
∴ y = 2
⇒ x cos ω + y sin ω = 2
∴ ω = \(\frac{\pi}{2}\), p = 2

(iii) x – y = 4
Put r cos ω = 1, r sin ω = – 1
r = \(\sqrt{1^{2}+(-1)^{2}}=\sqrt{2}\)
cos ω = \(\frac{1}{\sqrt{2}}\), sin ω = – \(\frac{1}{\sqrt{2}}\)
ω = 360° – 45°= 315°
Dividing eq. (i) by √2
\(\frac{1}{\sqrt{2}}\) x + (- \(\frac{1}{\sqrt{2}}\) y) = \(\frac{4}{\sqrt{2}}\) = 2√2
Normal form of the given line
x cos 315 + y sin 315 = 2√2
ω = 315°, p = 2√2.

Question 4.
Find the distance of the point (- 1, 1) from the line 12 (x + 6) = 5 (y – 2).
Answer.
The given equation of the line is 12(x + 6) = 5 (y – 2).
⇒ 12x + 72 = 5y – 10.
⇒ 12x – 5y + 82 = 0 ………………(i)
On comparing equation (i) with general equation of line Ax + By + C = 0, we obtain A = 12, B = – 5, and C = 82.
It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by
d = \(\frac{\left|A x_{1}+B y_{1}+C\right|}{\sqrt{A^{2}+B^{2}}}\)
The given point is (x₁, y₁) = (- 1, 1).
Therefore the distance of point (- 1, 1) from the given line
= \(\frac{|12(-1)+(-5)(1)+82|}{\sqrt{(12)^{2}+(-5)^{2}}}\) units

= \(\frac{|-12-5+82|}{\sqrt{169}}\) units

= \(\frac{|65|}{13}\) units = 5 units.

Question 5.
Find the points on the x-axis, whose distances from the line \(\) = 1 are 4 units.
Answer.
The given equation of line is \(\) = 1
or 4x + 3y -12 = 0 …………………(i)
On comparing equation (i) with general equation of line Ax + By + C = 0, we obtain A = 4, B = 3 and C = – 12.
Let (a, 0) be.the point on the x-axis whose distance from the given line is 4 units.
It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by

d = \(\frac{\left|A x_{1}+B y_{1}+C\right|}{\sqrt{A^{2}+B^{2}}}\)

Therefore, 4 = \(\frac{|4 a+3 \times 0-12|}{\sqrt{4^{2}+3^{2}}}\)

4 = \(\frac{|4 a-12|}{5}\)

⇒ |4a – 12| = 20
⇒ (4a – 12) = 20 or ± (4a – 12) = 20
⇒ (4a – 12) = 20 or – (4a – 12) = 20
⇒ 4a = 20 + 12 or 4a = – 20 + 12
⇒ a = 8or – 2
Thus, the required points on the x-axis are (- 2, 0) and (8, 0).

Question 6.
Find the distance between parallel lines
(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0
(ii) l(x + y) + p = 0 and l(x + y) – r = 0
Answer.
It is known that the distance (d) between parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0 is given by d = \(\frac{\left|C_{1}-C_{2}\right|}{\sqrt{A^{2}+B^{2}}}\)

(i) The given parallel lines are 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0.
Here, A = 15, B = 8, C₁ = – 34 and C₂ = 31.
Therefore, the distance between the piralle1 lines is
d = \(\frac{\left|C_{1}-C_{2}\right|}{\sqrt{A^{2}+B^{2}}}\)

= \(\frac{|-34-31|}{\sqrt{(15)^{2}+(8)^{2}}}\) units

= \(\frac{|-65|}{17}\) units = \(\frac{65}{17}\) units

(ii) The given parallel lines are l(x + y) + p =0 and l(x + y) – r = 0.
lx + ly + p = 0 and lx + ly – r = 0
Here, A = l, B = l, C₁ = p and C₂ = – r.
Therefore, the distance between the parallel lines is

d = \(\frac{\left|C_{1}-C_{2}\right|}{\sqrt{A^{2}+B^{2}}}\)

= \(\frac{|p+r|}{\sqrt{l^{2}+l^{2}}}\) units

= \(\frac{|p+r|}{\sqrt{2 l^{2}}}\) units

= \(\frac{|p+r|}{l \sqrt{2}}\) units

= \(\frac{1}{\sqrt{2}}\left|\frac{p+r}{l}\right|\) units.

Question 7.
Find equation of the line parallel to the line 3x – 4y + 2 = 0 and passing through the point (- 2, 3).
Answer.
The equation of the given line is 3x – 4y + 2 = 0
y = \(\)

or y = \(\), which is of the form y = mx + c
∴ Slope of the given line = \(\frac{3}{4}\)
It is known that parallel lines have the same slope.
∴ Slope of the other line = m = \(\frac{3}{4}\)
Now, the equation of the line that has a slope of \(\frac{3}{4}\) and passes through the point (- 2, 3) is
(y – 3) = \(\frac{3}{4}\) {x – (- 2)}
4y – 12 = 3x + 6
i.e., 3x – 4y + 18 = 0.

Question 8.
Find equation of the line perpendicular to the line x ly + 5=0 and having x intercept 3.
Answer.
Given equation of line is
x – 7y + 5 = 0 …………..(i)
The slope of a line is m₁ = \(\frac{1}{7}\)
∴ The slope of a perpendicular line is m = – 7.
Let the required perpendicular line be y = mx + c
y = – 7x + c [∵ m = – 7] ……………(i)
Since, this line intercept the X-axis at 3.
∴ Eq. (i) passes through point (3, 0).
∴ 0 = – 7(3) + c
⇒ c = 21
On putting c = 21 in eq. (i),we get
y = – 7x + 21
7x + y = 21

Question 9.
Find angles between the lines √3x + y = 1 and x + √3y =1.
Answer.
Given lines be √3x + y = 1 or y = – √3x + 1 …………(i)
and x + √3y = 1
or y = \(-\frac{1}{\sqrt{3}} x-\frac{1}{\sqrt{3}}\) ……………(ii)
∴ Slope of line (i) = – √3 = m₁ (say)
and Slope of line (ii) = – \(\frac{1}{\sqrt{3}}\) = m₂ (say)
If θ is the angle between be lines (i) and (ii) then

tan θ = \(\frac{\left|m_{1}-m_{2}\right|}{1+m_{1} m_{2}}=\frac{\left|-\sqrt{3}+\frac{1}{\sqrt{3}}\right|}{1+(-\sqrt{3}) \times\left(-\frac{1}{\sqrt{3}}\right)}\)

= \(\frac{\left|\frac{-3+1}{\sqrt{3}}\right|}{1+1}=\frac{2}{\sqrt{3}} \cdot \frac{1}{2}\)

= \(\frac{1}{\sqrt{3}}\) = tan \(\frac{\pi}{6}\)

θ = \(\frac{\pi}{6}\).

Question 10.
The line through the points (h, 3) and (4, 1) intersects the line 7x – 9y – 19 = 6, at right angle. Find the value of h.
Answer.
The slope of the line passing through points (h, 3) and (4, 1) is
m₁ = \(\frac{1-3}{4-h}=\frac{-2}{4-h}\)
The slope of line 7x – 9y – 19 = 0 or y = \(\frac{7}{9}\) x – \(\frac{19}{9}\) is m₂ = \(\frac{7}{9}\).
It is given that the two lines are perpendicular.
∴ m₁ × m₂ = – 1
⇒ 14 = 36 – 9h
⇒ 9h = 36 – 14
⇒ h = \(\frac{22}{9}\)
Thus, the value of h is \(\frac{22}{9}\).

Question 11.
Prove that the line through the point (x₁, y₁) and parallel to the line Ax + By + C = 0 is A(x – x) + B (y – y₁) = 0.
Solution:
The slope of line Ax + By + C = 0 or y = \(\left(\frac{-A}{B}\right) x+\left(\frac{-C}{B}\right)\) is m = \(-\frac{A}{B}\)
It is known that parallel lines have the same slope.
∴ Slope of the other line = m = \(-\frac{A}{B}\)
The equation of the line passing through point (x₁, y₁) and having a slope m = \(-\frac{A}{B}\) is
y – y₁ = m (x – x₁) = \(-\frac{A}{B}\) (x – x₁)
B(y – y₁) = – A (x – x₁)
A(x – x₁) + B(y – y₁) = 0.
Hence, the line through point (x₁, y₁) and parallel to line Ax + By + C = 0 is
A(x – x₁) + B(y – y₁) = 0.

Question 12.
Two lines passing through the point (2, 3) intersects each other at an Rngle of 60°. If slope of one line is 2, find equation of the other line.
Answer.
Let m be the slope of the other line.
Slope of line = 2
Angle between them = 60°
tan 60° = ± \(\frac{m-2}{1+2 m}\)

\(\frac{m-2}{1+2 m}\) = ± √3

(+ve) sign, m – 2 = √3 (1 + 2m) = √3 + 2√3m
or (2√3 – 1 )m = – 2 – √3
m = – \(\frac{(2+\sqrt{3})}{2 \sqrt{3}-1}\)
Equation of the line passing through the point (2, 3) with slope m is
y – 3 = \(\frac{(2+\sqrt{3})}{2 \sqrt{3}-1}\) (x – 2)

⇒ (2√3 – 1) y – 3 (2√3 – 1) = – (2 + √3) x + 2 (2 + √3)
⇒ (2√3 – 1) y – 6√3 + 3 = – (2 + √3) x + 4 + 2√3
⇒ (2 + √3) x + (2√3 – 1) y – 6√3 + 3 – 4 – 2√3 = 0
⇒ (2 + √3) x + (2√3 – 1) y – 8√3 – 1 = 0

Taking negative, sign \(\frac{m-2}{1+2 m}\) = – √3
m – 2 = – √3 (1 + 2m) = – √3 – 2√3m
(2√3 + 1) m = 2 – √3
m = \(\frac{2-\sqrt{3}}{1+2 \sqrt{3}}\)

So, equation of the line passing through (2, 3) with slope \(\frac{2-\sqrt{3}}{1+2 \sqrt{3}}\) is

y – 3 = \(\frac{2-\sqrt{3}}{1+2 \sqrt{3}}\) (x – 2)
⇒ (1 + 2√3) y – 3 (1 + 2√3) = (2 – √3) x – 2 (2 – √3)
⇒ (1 + 2√3) y – 3 – 6√3 = (2 – √3) x – 4 + 2√3
⇒ (2 – √3)x – (1 + 2√3) y – 1 + 8 = 0
Hence, required lines are
(√3 + 2)x + (2√3 – 1)y – 8√3 – 1 = 0
and (2 – √3) x – (1 + 2√3) + 8√3 – 1 = 0.

Question 13.
Find the equation of the right bisector of the line segment joining the points (3, 4) and (- 1, 2).
Answer.
The right bisector of a line segment bisects the line segment at 90°.
The end-points of the line segment are given as A(3, 4) and B (- 1, 2).
Accordingly, mid – point of AB = \(\left(\frac{3-1}{2}, \frac{4+2}{2}\right)\) = (1, 3).
Slope of AB = \(\frac{2-4}{-1-3}=\frac{-2}{-4}=\frac{1}{2}\)
∴ Slope of the line perpendicular to AB = \(-\frac{1}{\left(\frac{1}{2}\right)}\) = – 2.
The equation of the line passing through (1, 3) and having a slope of – 2 is
(y – 3) = – 2 (x – 1)
y – 3 = – 2x + 2
2x + y = 5
Thus, the required ecuation of the line is 2x + y = 5.

Question 14.
Find the coordinates of the foot of perpendicular from the point (-1,3) to the line 3a; – 4y -16 = 0.
Answer.
The equation of the given line is 3x – 4y – 16 = 0
The equation of a line perpendicular to the given line is 4x + 3y + k = 0, where is a constant.
If this point passes through the point (- 1, 3) then
– 4 + 9 + k = 0
k = – 5
∴ The equation of a line passing through the point (- 1, 3) and perpendicular to the given line is
4x – 3y – 5 = 0
∴ The required foot of the perpendicular is the point of intersection of the lines
3x – 4y – 16 = 0 …………….(i)
and 4x + 3y – 5 = 0 …………….(ii)
Solving eqs. (i) and (ii) y cross – multiplication, we have

Hence, co-ordinates of first perpendicular are (\(\frac{68}{25}\), – \(\frac{49}{25}\)).

Question 15.
The perpendicular from the origin to the line y = mx + c meets it at the point (- 1, 2). Find the values of m and c.
Answer.
The given equation of line is y = mx + c;
It is given that the perpendicular from the origin meets the given line at (- 1, 2).
Therefore, the line joining the points (0, 0) and (- 1, 2) is perpendicular to the given line.
∴ Slope of the line joining (0, 0) and (- 1, 2) = \(\frac{2}{-1}\) = – 2
The slope of the given line is in.
∴ m × – 2 = – 1 [The two lines are perpendicular]
⇒ m = \(\frac{1}{2}\)
Since point (- 1, 2) lies on the given line, it satisfies the equation y = mx + c.
∴ 2 = m (- 1) + c
2 = \(\frac{1}{2}\) (- 1) + c
⇒ c = 2 + \(\frac{1}{2}\) = \(\frac{5}{2}\).
Thus, the respective values of m and c are \(\frac{1}{2}\) and \(\frac{5}{2}\).

Question 16.
If p and q are the lengths Qf perpendiculars from the origin to the lines x cos θ – y sin θ = k cos 2θ and x sec θ + y cosec θ = k respectively, prove that p² + 4q² = k².
Answer.
p = the length of perpendicular from origin (0, 0) to the line x cos θ – y sin θ – k cos 2θ = 0
Then, p = \(\frac{|0 \cdot \cos \theta+0 \cdot(-\sin \theta)-k \cos 2 \theta|}{\sqrt{\cos ^{2} \theta+\sin ^{2} \theta}}\)

[∵ the perpendicular distance from (x₁, y₁) tot the line ax + y + c = 0 is \(\frac{a x_{1}+b y_{1}+c}{\sqrt{a^{2}+b^{2}}}\)].

∴ p = \(\frac{k \cos 2 \theta}{1}\)
⇒ p = k cos 2θ ………………….(i)

and q = the length of perpendicular from (0, 0) to the line x sec θ + y cosec θ – k = 0
Then, q = \(\frac{|0 \cdot \sec \theta+0 \cdot \ {cosec} \theta-k|}{\sqrt{\sec ^{2} \theta+\ {cosec}^{2} \theta}}\)

q = \(\frac{k}{\sqrt{\frac{1}{\cos ^{2} \theta}+\frac{1}{\sin ^{2} \theta}}}=\frac{k}{\sqrt{\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin ^{2} \theta \cdot \cos ^{2} \theta}}}\)

= \(\frac{k \cdot \sin \theta \cos \theta}{1} \times \frac{2}{2}\)
[∵ sin² θ + cos² θ = 1]

q = \(\frac{k}{2}\) sin 2θ
2q = k sin 2θ ……………(ii)

On squaring eqs. (i) and (ii) and then adding, we get
p² + 4q² = k² cos² 2θ + k² sin² 2θ
= k² (cos² 2θ + sin² 2θ)
p² + 4q² = k²
[∵ cos² θ + sin² θ = 1]
Hence proved.

Question 17.
In the triangle ABC with vertices A(2, 3), B(4, 1) and C(1, 2), fInd the equation and length of altitude from the vertex A.
Answer.

Let AD be the altitude of triangle ABC from vertex A.
Accordingly, AD ⊥ BC
The equation of the line passing through point (2, 3) and having a slope of 1 is
(y – 3) = 1(x – 2)
⇒ x – y + 1 = 0
⇒ y – x = 1
Therefore, equation of the altitude from vertex A is y – x = 1.
Length of AD = Length of the perpendicular from A(2, 3) to BC.
The equation of BC is (y + 1) = \(\frac{2+1}{1-4}\) (x – 4)
⇒ (y + 1) = – 1 (x – 4)
⇒ y + 1 = – x + 4
⇒ x + y – 3 = 0 ……………… (i)
The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by
d = \(\frac{\left|A x_{1}+B y_{1}+C\right|}{\sqrt{A^{2}+B^{2}}}\)
On comparing equation (i) to the general equation of line Ax + By + C = 0,
we obtain A = 1, B = 1 and C = – 3.
∴ Length of AD = \(\frac{|1 \times 2+1 \times 3-3|}{\sqrt{1^{2}+1^{2}}}\) units
= \(\frac{|2|}{\sqrt{2}}\) units

= \(\frac{2}{\sqrt{2}}\) units

= √2 units.

Thus, the equation and the length of the altitude from vertex A are y – x = 1 and √2 units respectively.

Question 18.
If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that
\(\frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}\).
Answer.
It is known that the equatìon of a line whose intercepts on the axes are a and b is
\(\frac{x}{a}+\frac{y}{b}\) = 1
or bx + ay = ab or bx + ay – ab = 0 …………….(i)
The perpendicular distance (d) of a line Ax + By + C = O from a point(x₁, y₁) is given by
d = \(\frac{\left|A x_{1}+B y_{1}+C\right|}{\sqrt{A^{2}+B^{2}}}\)
On comparing equation (i) to the general equation of line Ax + By + C = 0, we obtain A = b, B = a, and C = – ab.
Therefore, if p is the length of the perpendicular from point (x₁, y₁) = (0, 0) to line (i), we obtain

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 10 Straight Lines Ex 10.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.2

Question 1.
Write the equations for the x and y-axes.
Answer.
The y-coordinate of every point on the x-axis is 0.
Therefore, the equation of the x – axis is y = 0.
The x-coordinate of every point on the y – axis is 0.
Therefore, the equation of the y-axis is x = 0.

Question 2.
Find the equation of the line which passes through the point (- 4, 3) with slope \(\frac{1}{2}\).
Answer.
We know that the equation of the line passing through point (x₀, y₀), whose slope is m, is (y – y₀) = m (x – x₀).
Thus, the equation of the line passing through point (- 4, 3), whose slope is \(\frac{1}{2}\) is
(y – 3) = – (x + 4)
2 (y – 3) = x + 4
2y – 6 = x + 4
i.e., x – 2y + 10 = 0.

Question 3.
Find the equation of the line which passes through (0, 0) with slope m.
Answer.
We know that the equation of the line passing through point (x₀, y₀), whose slope is m, is
(y – y₀) = m (x – x₀)
Thus, the equation of the line passing through point (0, 0), whose slope is m is (y – 0) = m(x – 0)
i.e., y = mx.

Question 4.
Find the equation of the line which passes through (2, 2√3) and is inclined with the x-s’ at an angle of 75°.
Answer.
The slope of the line that inclines with the x-axis at an angle of 75° is m = tan75°
m = tan (45° + 30°)
= \(\frac{\tan 45^{\circ}+\tan 30^{\circ}}{1-\tan 45^{\circ} \tan 30^{\circ}}\)

= \(\frac{1+\frac{1}{\sqrt{3}}}{1-1 \cdot \frac{1}{\sqrt{3}}}\)

= \(\frac{\frac{\sqrt{3}+1}{\sqrt{3}}}{\frac{\sqrt{3}-1}{\sqrt{3}}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}\)

We know that the equation of the line passing through point (x₀, y₀), whose slope is m, is (y – y₀) = m(x – x₀).
Thus, if a line passes though (2, 2√3) and inclines with the x-axis at an angle of 75°, then the equation of the line is given as
(y – 2√3) = (x – 2)
(y – 2√3) (√3 – 1) = (√3 + 1)(x – 2)
y(√3 – 1) – 2 √3 (√3 – 1) = x(√3 + 1) – 2 (√3 + 1)
(√3 + 1) x – (√3 – 1) y = 2√3 + 2 – 6 + 2√3
(√3 + 1) x – (√3 – 1) y = 4√3 – 4
i.e., (√3 + 1) x – (√3 – 1) y = 4(√3 – 1).

Question 5.
Find the equation of the line which intersects the x-axis at a distance of 3 units to the left of origin with slope – 2.
Answer.
It is known that if a line with slope m makes x-intercept d, then the equation of the line is given as y = m(x – d).
For the line intersecting the x-axis at a distance of 3 units to the left of the origin, d = – 3.
The slope of the line is given as m = – 2.
Thus, the required equation of the given line is
y = – 2 [x – (- 3)] = – 2x – 6
i.e., 2x + y + 6 = 0.

Question 6.
Find the equation of the line which intersects the y-axis at a distance of 2 units above the origin and makes an angle of 30° with the positive direction of the x-axis.
Answer.

The line intersect the 7-axis at distance of 2 units above the origin. It show that the line passes through (0, 2).
It makes an angle 30° with the positive direction of X – axis.
1 X
So, the slope of the line is = tan 30° = \(\frac{1}{\sqrt{3}}\)
Thus, the equation of straight line is
y – y₁ = m (x – x₁)
y – 2 = \(\frac{1}{\sqrt{3}}\) (x – 0)
\(\frac{x}{\sqrt{3}}\) – y + 2 = 0
x – √3y + 2√3 = 0.

Question 7.
Find the equation of the line which passes through the points (- 1, 1) and (2, – 4).
Answer.
Given points are A (x₁, y₁) = (- 1, 1) and B(x₂, y₂) = (2, – 4), then equation of line AB is
y – y₁ = \(\frac{y-{2}-y-{1}}{x-{2}-x-{1}}\) (x – x₁)
⇒ y – 1 = \(\frac{-4-1}{2+1}\) (x + 1)
[x₁ = – 1, y₁ = 1, x₂ = 2, y₂ = – 4]
⇒ y – 1 = – \(\frac{5}{3}\) (x + 1)
⇒ 3y – 3 = – 5x – 5
⇒ 5x + 3y + 2 = 0

Question 8.
Find the equation of the line which is at a perpendicular distance of 5 units from the origin and the angle made by the perpendicular with the positive x – axis is 30°.
Answer.
If p is the length of the normal from the origin to a line and A is the angle made by the normal with the positive direction of the x-axis, then the equation of the line is given by x cos A + y sin A = p
Here, p = 5 units and A = 30°
Thus, the required equation of the given line is x cos 30° + y sin 30° = 5
x \(\frac{\sqrt{3}}{2}\) + y . \(\frac{1}{2}\) = 5
i.e., √3x + y = 10.

Question 9.
The vertices of ∆PQR are P(2, 1), Q(- 2, 3) and R(4, 5). Find equation of the median through the vertex R.
Answer.
Since, median bisects the opposite sides i. e., S is the mid-point of PQ.
S = \(\left(\frac{\ddot{x}-{1}+x-{2}}{2}, \frac{y-{1}+y-{2}}{2}\right)\)

= \(\left(\frac{2-2}{2}, \frac{1+3}{2}\right)=\left(0, \frac{4}{2}\right)\)

= (0, 2)

y – y₁ = \(\frac{y-{2}-y-{1}}{x-{2}-x-{1}}\) (x – x₁)
⇒ y – 5 = \(\frac{2-5}{0-4}\) (x – 4)
[∵ x₁ = 2, y₁ = 1, x₁ = – 2, y₁ = 3]
⇒ y – 5 = \(\frac{-3}{-4}\) (x – 4)
4y – 20 = 3x – 12
⇒ 3x – 4y + 8 = 0

Question 10.
Find the equation of the line passing through (- 3, 5) and perpendicular to the line through the points (2, 5) and (- 3, 6).
Answer.
The slope of the line joining the points (2, 5) and (- 3, 6) is
m = \(\frac{6-5}{-3-2}=\frac{1}{-5}\)

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.

Therefore, slope of the line perpendicular to the line through the points (2, 5) and (- 3, 6)
= \(-\frac{1}{m}=-\frac{1}{\left(\frac{-1}{5}\right)}\) = 5

Now, the equation of the line passing through point (- 3, 5), whose slope is 5, is
(y – 5) = 5 (x + 3)
y – 5 = 5x + 15
i.e., 5x – y + 20 = 0.

Question 11.
A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1 : n. Find the equation of the line.
Answer.

Let the given points are A(1, 0) and B(2, 3).
Let the line PQ divide AB in the ratio 1 : n at R internally.
Then, coordinates of R = \(\left(\frac{1 \times x-{2}+n \times x-{1}}{1+n}, \frac{1 \times y-{2}+n \times y-{1}}{1+n}\right)\)

= \(\left(\frac{1 \times 2+n \times 1}{1+n}, \frac{1 \times 3+n \times 0}{1+n}\right)\)

[∵ x₁ = 1, y₁ = 0, x₂ = 2, y₂ = 3]

= \(\left(\frac{n+2}{n+1}, \frac{3}{1+n}\right)\)

Let the slope of the line is m.
Also, PQ ⊥ AB
∴ Slope of the line PQ × Slope of the line AB = – 1
⇒ m × \(\frac{y-{2}-y-{1}}{x-{2}-x-{1}}\) = – 1

⇒ m × \(\frac{3-0}{2-1}\) = – 1

[∵ x₁ = 1, y₁ = 0, x₂ = 2, y₂ = 2]

⇒ m × 3 = – 1
⇒ m = – \(\frac{1}{3}\)

Now, equation of line PQ y using y – y₀ = m (x – x₀)

y – \(\frac{3}{1+n}\) = \(\frac{-1}{3}\left(x-\frac{n+2}{n+1}\right)\)

[∵ R\(\left(\frac{n+2}{n+1}, \frac{3}{1+n}\right)\) = (x₁, y₁)

\(\frac{3(n+1) y-9}{1+n}=\frac{-x(n+1)+(n+2)}{n+1}\)

⇒ 3 (n + 1) y – 9 = – x (n + 1) + (n + 2)
⇒ x (n + 1) + 3 (n + 1) y = n + 2 + 9
⇒ x (n + 1) + 3 (n + 1) y = n + 11
Which is the required equation of line.

Question 12.
Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).
Answer.
The equation of a line in the intercept form is
\(\frac{x}{a}+\frac{y}{b}\) = 1 …………….(i)
Here, a and b are the intercepts on x and y axes respectively.
It is given that the line cuts off equal intercepts on both the axes.
This means that a = b.
Accordingly, equation (i) reduces to
\(\frac{x}{a}+\frac{y}{a}\) = 1
x + y = a ………………(ii)
Since the given line passes through point (2, 3), equation (ii) reduces to
2 + 3 = a
a = 5
On substituting the value of a in equation (ii), we obtain x + y = 5, which is the required equation of the line.

Question 13.
Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.
Answer.
The equation of a line in the intercept form is
\(\frac{x}{a}+\frac{y}{b}\) = 1 ……………(i)
Here, a and b are the intercepts on x and y axes respectively.
It is given that a + b = 9
⇒ b = 9 – a ………….. (ii)
From equations (i) and (ii), we obtain
\(\frac{x}{a}+\frac{y}{9-a}\) = 1 ………..(iii)

It is given that the line passes through point (2, 2).
Therefore, equation (iii) reduces to \(\frac{2}{a}+\frac{2}{9-a}\) = 1
⇒ \(2\left(\frac{1}{a}+\frac{1}{9-a}\right)\) = 1

⇒ \(2\left(\frac{9-a+a}{a(9-a)}\right)\) = 1

⇒ \(\frac{18}{9 a-a^{2}}\) = 1

⇒ 18 = 9a – a²
⇒ a² – 9a + 18 = 0
⇒ a² – 6a – 3a + 18 = 0
⇒ a (a – 6) – 3 (a – 6) = 0
⇒ (a – 6) (a – 3) = 0
⇒ a = 6 or a = 3
If a = 6 and b = 9 – 6 =3, then the equation of the line is \(\frac{x}{6}+\frac{y}{3}\) = 1
⇒ x + 2y – 6 = 0
If a = 3 and = 9 – 3 = 6, then the equation of the line is
\(\frac{x}{6}+\frac{y}{3}\) = 1
⇒ 2x + y – 6 = 0.

Question 14.
Find equation of the line through the point (0, 2) making an angle – with the positive x-axis. Also, find the equation of line 3 parallel to it and crossing the y-axis at a distance of 2 units below the origin.
Answer.

Slope of the line = tan \(\frac{2 \pi}{3}\)
tan (π – \(\frac{\pi}{3}\)) = tan \(\frac{\pi}{3}\) = – √3
Equation of the line AC with slope = – √3 and passing through the point D given by
y – 2 = – √3(x – 0)
⇒ √3x + y – 2 = 0
For beyond past, we take another line BD parallel to AC.
∴ Slope of the line BD = Slope of AC = – √3
BD is passing through the point D{0, – 2)
∴ Equation of BD is y + 2 = √3(x – 0)
⇒ √3x + y + 2 = 0
Hence, equation of AC and BD are,
√3x + y – 2 = 0 and √3x + y + 2 = 0.

Question 15.
The perpendicular from the origin to a line meets it at the point (- 2, 9), find the equation of the line.
Answer.
The slope of the line joining the origin (0, 0) and point (- 2, 9) is
m₁ = \(\frac{9-0}{-2-0}=-\frac{9}{2}\)

Accordingly, the slope of the line, perpendicular to the line joining the origin and point (- 2, 9) is
m₂ = \(-\frac{1}{m-{1}}=-\frac{1}{\left(-\frac{9}{2}\right)}=\frac{2}{9}\)

Now, the equation of the line passing through point (- 2, 9) and having a slope m2 is
(y – 9) = \(\frac{2}{9}\) (x + 2)
9y – 81 = 2x + 4
2x – 9y + 85 = 0.

Question 16.
The length L (in centimetre) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L = 124.942 when C = 20 and L = 125.134 when C = 110, express L in terms of C.
Answer.
Since, L is linear function of C.
L = a + bC …………….(i)
For L = 124.942, C = 20
124.942 = a + 20b ………………(ii)
For L = 125.134 C = 110
∴ 125.134 = a + 1106 ……………(iii)
Subtracting eq. (ii) from eq. (iii), we get
0. 192 = 90b
b = \(\frac{0.192}{90}\) = 0.00213
From eq. (ii),
124.942 = a + 20 × 0.00213 = a + 0.0426
a = 124.942 – 0.0426 = 124.8994
Hence from eq. (i) L in terms of C is
L = 124.8994 + 0.00213C.

Question 17.
The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs. 14/litre and 1220 litres of milk each week at Rs. 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs. 17/litre?
Answer.
Let price and litre be denoted in ordered pair (x, y), where x denotes the ₹ per litre and y denotes the quantity of milk in litre.
Given, (14980) and (161220) are two points.
Let linear relations i. e., linear equation be
y – y₁ = \(\frac{y-{2}-y-{1}}{x-{2}-x-{1}}\) (x – x₁)

y – 980 = \(\frac{1220-980}{16-14}\) (x – 14)

y – 980 = \(\frac{240}{2}\) (x – 14)
[∵ x₁ = 14, y₁ = 980, x₂ = 16, y₂ = 1220]
⇒ y – 980 = 120 (x – 14)
⇒ y – 980 = 120x – 120 × 14
⇒ 120x – y = 1680 – 980
⇒ 120x – y = 700
When price JC = 17,
120 x 17 – y = 700
y = 2040 – 700 = 1340
He will sell weekly 1340 L milk at the rate of ₹ 172.

Question 18.
P(a, b) is the mid-point of a line segment between axes. Show that equation of the line is \(\frac{x}{a}+\frac{y}{b}\) = 2.
Answer.
Let AB be the line segment between the axes and let P(a, b) be its mid-point.

Let the coordinates of A and B be (0, y) and (x, 0) respectively.
Since P(a, b) is the mid-point of AB.
\(\left(\frac{0+x}{2}, \frac{y+0}{2}\right)\) = (a, b)

\(\left(\frac{x}{2}, \frac{y}{2}\right)\) = (a, b) and

\(\frac{x}{2}\) = a and \(\frac{y}{2}\) = b
x = 2a and y = 2b
Thus, the respectively coordinares of A and B are (0, 2b) and (2a, 0).
The equation of the line passing through points (0, 2b) and (2a, 0) is

(y – 2b) = \(=\frac{(0-2 b)}{(2 a-0)}\) (x – 0)

y – 2b = \(\frac{-2 b}{2 a}\) (x)

a (y – 2b) = – bx
ay – 2ab = – bx
i.e., bx + ay = 2ab
On dividing both sides by ab, we obtain
\(\frac{b x}{a b}+\frac{a y}{a b}=\frac{2 a b}{a b}\)

⇒ \(\frac{x}{a}+\frac{y}{b}\) = 2
Thus, the equation of the line is \(\frac{x}{a}+\frac{y}{b}\) = 2.

Question 19.
Point R (h, k) divides a line segment between the axes in the ratio 1 : 2. Find equation of the line.
Answer.

Let AB be the line segment between the axes such that point R(h, k) divides AB in the ratio 1 : 2.
Let the respective coordinates of A and B be (x, 0) and (0, y).
Since point R(h, k) divides AB in the ratio 1 : 2, according to the section formula,
(h, k) = \(=\left(\frac{1 \times 0+2 \times x}{1+2}, \frac{1 \times y+2 \times 0}{1+2}\right)\)

(h, k) = \(\left(\frac{2 x}{3}, \frac{y}{3}\right)\)

⇒ h = \(\frac{2 x}{3}\) and k = \(\frac{y}{3}\)
⇒ x = \(\frac{3 h}{2}\) and y = 3k

Therefore, the respective coordinates of A and B are (\(\frac{3 h}{2}\), o) and (0, 3k).

Now, the equation of line AB passing through points (\(\frac{3 h}{2}\), 0) and (0, 3k) is

(y – 0) = \(\frac{3 k-0}{0-\frac{3 h}{2}}\) (x – \(\frac{3 h}{2}\))

y = – \(\frac{2 k}{h}\) (x – \(\frac{3 h}{2}\))
hy = – 2kx + 3hk
i.e., 2kx + hy = 3hk
Thus, the required equation of the line is 2kx + hy = 3hk.

Question 20.
By using the concept of equation of a line, prove that the three points (3, 0), (- 2, – 2) and (8, 2) are collinear.
Answer.
In order to show that points (3, 0) (- 2, – 2) and (8, 2) are coimear, it suffices to show that the line passing through points (3, 0) and (- 2, – 2) also passes through point (8, 2).
The equation of the line passing through points (3, 0) and (- 2, – 2) is
(y – 0) = \(\frac{(-2-0)}{(-2-3)}\) (x – 3)

y = \(\frac{-2}{-5}\) (x – 3)
5y = 2x – 6
i.e., 2x – 5y = 6
It is observed that at x = 8 and y = 2,
5y = 2x – 6
i.e., 2x – 5y = 6
It is observed that at x = 8 and y = 2,
L.H.S. = 2 × 8 – 5 × 2
= 16 – 10 = 6 = R.H.S.
Therefore, the line passing through points (3, 0) and (- 2, – 2) also passes through point (8, 2).
Hence, points (3, 0), (- 2, – 2) and (8, 2) are collinear.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 10 Straight Lines Ex 10.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.1

Question 1.
Draw a quadrilateral in the Cartesian plane, whose vertices are (- 4, 5), (0, 7), (5, – 5) and (- 4, – 2). Also find its area.
Answer.
Let ABCD be the given quadrilateral with vertices A (- 4, 5), B (0, 7), C (5 – 5), and D (- 4. – 2).
Then, by plotting A, B, C and D on the Cartesian plane and joining AB, BC, CD and DA the given quadrilateral can be drawn as

To find the area of quadrilateral ABCD, we draw one diagonal, say AC.
Accordingly, area (ABCD) = area (∆ABC) + area (∆ACD)
We know that the area of a triangle whose vertices are (x₁, y₁), (x₂, y₂) and (x₃, y₃) is \(\frac{1}{2}\) [x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]
Therefore, area of ∆ABC
= \(\frac{1}{2}\) |- 4 (7 + 5) + 0(- 5 – 5) + 5 (5 – 7)| unit²
= \(\frac{1}{2}\) |- 4 (12) + 5 (- 2)| unit²
= \(\frac{1}{2}\) |- 48 – 10| unit²
= \(\frac{1}{2}\) |- 58| unit²
= \(\frac{1}{2}\) × 58 unit²
= 29 unit²
Area of MCD = \(\frac{1}{2}\) |- 4 (- 5 + 2) + 5(- 2 – 5) ± (- 4) (5 + 5)| unit²
= \(\frac{1}{2}\) |- 4 (- 3) + 5 (- 7) – 4(10)| unit²
= \(\frac{1}{2}\) |12 – 35 – 40| unit²
= \(\frac{1}{2}\) |- 63| unit²
= \(\frac{63}{2}\) unit²
Thus, area (ABCD) = (29 + \(\frac{63}{2}\)) unit²
= \(\frac{58+63}{2}\) unit²
= \(\frac{121}{2}\) unit².

Question 2.
The base of an equilateral triangle with side 2a lies along the y-axis such that the mid point of the base is at the origin. Find vertices of the triangle.
Answer.

Let BC be the base of a triangle which lies on Y
Y-axis and third vertex may be A(h, 0) or A’.
Since, ∆ABC is an equilateral, then AB = BC.
∴ AB² = BC²
⇒ (h – 0)² + (0 – a²) = (2a)²
[∵ distance between two points (x₁, y₁) and (x₂, y₂)
= \(\left.\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\right]\)
For distance AB, (x₁, y₁) = (a, 0), (x₂, y₂) = (0, a)
h² + a² = 4a²
h² = 3a²
h = ± √3 a [taking square root]
Hence, the vertices of triangle are (√3a, 0), (0, a), (0,- a) or (- √3, a) (0, a), (0, – a).

Question 3.
Find the distance between P(x₁, y₁) and Q(x₂, y₂) when:
(i) PQ is parallel to the y – axis
(ii) PQ is parallel to the x-aLg.
Answer.

(i) When PQ is parallel to the Y-axis, it means the x-coordinates of P and Q are same i.e.,
x₁ = x₂.
∴ Distance between two points
PQ = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
= \(\sqrt{\left(x_{1}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
= \(\sqrt{\left(y_{2}-y_{1}\right)^{2}}=\left|y_{2}-y_{1}\right|\)

(ii)

When PQ is parallel to X-axis, it means y-coordinates of P and Q are same i.e., y₂ = y₁
∴ Distance between two points PQ = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)

= \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{1}-y_{1}\right)^{2}}\) [∵ y₂ = y₁]

= \(\sqrt{\left(x_{2}-x_{1}\right)^{2}}\)

= |x₂ – x₁|

Question 4.
Find a point on the x-axis, which is equidistance from the points (7, 6) and (3, 4).
Answer.
Let any point P on the X-axis is (x, 0) as for a point on X-axis y-coordinate is zero and the given points are A (7, 6) and B (3, 4).
Given, PA = PB
⇒ PA² = PB²
⇒ (x₁ – x₂)² + (y₁ – y₂)² = (x₁ – x₃)² + (y₁ – y₃)²
where x₁ = x, x₂ = 7, y₁ = 0, y₂ = 6, x₃ = 3, y₃ = 4
(x – 7)² + (0 – 6)² = (x – 3)² + (0 – 4)²
[∵ distance between two points = \(\left.\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}\right]\)]
x² + 49 – 14x + 36 = x² + 9 – 6x + 16
⇒ – 14x + 6x = 25 – 36 – 49
⇒ – 8x = 25 – 85
⇒ Point P on X-axis = (\(\frac{15}{2}\), 0).

Question 5.
Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P(0, – 4) and B(8, 0).
Answer.

Given points are p (0, – 4) and Q (8, 0).
∴ x₁ = 0, y₁ = – 4, x₂ = 8, y₂ = 0
These points plotted in XY-plane are given below.
Mid-point of PQ is R = \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\)

= \(\left(\frac{0+8}{2}, \frac{-4+0}{2}\right)\) = (4, – 2)

∴ Slope of the OR = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

= \(\frac{-2-0}{4-0}=\frac{-2}{4}=-\frac{1}{2}\)

[∵ x = 0, x₂ = 4, y₁ = 0, y₂ = – 2]

Question 6.
Without using the Pythagoras theorem show that the points (4, 4), (3, 5) and (- 1, – 1) are the vertices of a right angled triangle.
Answer.
In ∆ABC, we have
m₁ = Slope of AB = \(\frac{4-5}{4-3}\) = – 1 and
m₂ = Slope of AC = \(\frac{4-(-1)}{4-(-1)}\) = 1
Clearly, m₁ m₂ = – 1
This shows that AB is perpendicular to AC i. e.,
∠CAB = π/2
Hence, the given points are the vertices of a right-angled triangle.

Question 7.
Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.
Answer.
The line OP makes an angle of 30° with y-axis measured anticlockwise.

So OP makes an angle of 90° + 30° = 120° with positive direction of x-axis.
So, slope of OP = tan 120° = tan (180° – 60°) = – tan 60° = – √3.

Question 8.
Find the value of x for which the points (x, – 1), (2, 1) and (4, 5) are collinear.
Answer.
Given, points A (x, – 1), B (2, 1) and C (4, 5) are collinear.
Here, x₁ = x, y₁ = – 1, x₂ = 2, y₂ = 1, x₃ = 4 and y₃ = 5
∴ Slope of AB = Slope of BC
⇒ \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{y_{3}-y_{2}}{x_{3}-x_{2}}\)

⇒ \(\frac{1+1}{2-x}=\frac{5-1}{4-2}\)

⇒ \(\frac{2}{2-x}=\frac{4}{2}\)

⇒ 2 – x = 1
⇒ x = 2 – 1 = 1

Question 9.
Without using distance formula, show that points (- 2, – 1), (4, 0), (3, 3) and (- 3, 2) are vertices of a parallelogram.
Answer.
Let ABCD be a parallelogram, where vertices are

(x₁, y₁) → A (- 2, – 1),
(x₂, y₂) → B (4, 0),
(x₃, y₃) → C (3, 3) and
(x₄, y₄) → D (- 3, 2)
Mid-point of AC = \(\left(\frac{x_{1}+x_{3}}{2}, \frac{y_{1}+y_{3}}{2}\right)=\left(\frac{-2+3}{2}, \frac{-1+3}{2}\right)\)

= \(\left(\frac{1}{2}, \frac{2}{2}\right)=\left(\frac{1}{2}, 1\right)\)

[∵ mid-point of two points = \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\)]

Mid point of BD = \(\left(\frac{x_{2}+x_{4}}{2}, \frac{y_{2}+y_{4}}{2}\right)\)

= \(\left(\frac{4-3}{2}, \frac{0+2}{2}\right)=\left(\frac{1}{2}, 1\right)\)

∵ Mid-point of AC = Mid-point of BD

∴ ABCD is parallelogram.
Hence proved.

Question 10.
Find the angle between the xr-axis and the line joining the points (3, – 1) and (4, – 2).
Answer.
The slope of the line joining the points (3, – 1) and (4, – 2) is
m = \(\frac{-2-(-1)}{4-3}\)
= – 2 + 1 = – 1
Now, the inclination (θ) of the line joining the points (3, – 1) and (4, – 2) is given by tan θ = – 1
⇒ θ = (90° + 45°) = 135°
Thus, the angle between the x-axis and the line joining the points (3, – 1) and (4, – 2) is 135°.

Question 11.
The slope of a line is double of the slope of another line. If tangent of the angle between them is \(\frac{1}{3}\), find the slopes of the lines.
Answer.
Let m₁ and m be the slopes of the two given lines such that m₁ = 2m
We know that if 0 is the angle between the lines l₁ and l₂ with slopes m₁ and m₂ then
tan θ = \(\left|\frac{m_{2}-m_{1}}{1+m_{1} m_{2}}\right|\)
It is given that the tangent of the angle between the two lines is \(\frac{1}{3}\).
∴ \(\frac{1}{3}=\left|\frac{m-2 m}{1+(2 m) \cdot m}\right|\)

\(\frac{1}{3}=\left|\frac{-m}{1+2 m^{2}}\right|\)

\(\frac{1}{3}=\frac{-m}{1+2 m^{2}}\) or

\(\frac{1}{3}=-\left(\frac{-m}{1+2 m^{2}}\right)=\frac{m}{1+2 m^{2}}\)

Case I:

\(\frac{1}{3}=\frac{-m}{1+2 m^{2}}\)

⇒ 1 + 2 m² = – 3m
⇒ 2m + 3m + 1 = 0
⇒ 2m² + 2m + m + 1 = 0
⇒ (m + 1) (2m + 1) = 0
⇒ m = – 1 or m = – \(\frac{1}{2}\)
If m = – 1, then the slopes of the lines are – 1 and – 2.
If m = – \(\frac{1}{2}\), then the slopes of the lines are – \(\frac{1}{2}\) and – 1.

Case II:
\(\frac{1}{3}=\frac{m}{1+2 m^{2}}\)
⇒ 2m² + 1 = 3m
⇒ 2m² – 2m – m + 1 = 0
⇒ (m – 1) (2m – 1) = 0
⇒ m = 1 or m = \(\frac{1}{2}\)
If m = 1, then the slopes of the lines are 1 and 2.
If m = \(\frac{1}{2}\), then the slopes of the lines are \(\frac{1}{2}\) and 1.
Hence, the slopes of the lines are – 1 and – 2 or – \(\frac{1}{2}\) and – 1 or 1 and 2 or \(\frac{1}{2}\) and 1.

Question 12.
A line passes through (x₁, y₁) and (h, k). If slope of the line is m, show that k – y₁ = m (h – x₁).
Answer.
The slope of the line passing through (x₁, y₁) and (h, k) is \(\frac{k-y_{1}}{h-x_{1}}\).

It is given that the slope of the line is

∴ \(\frac{k-y_{1}}{h-x_{1}}\) = m
⇒ k – y₁ = m (h – x₁)
Hence, k – y₁ = m (h – x₁).

Question 13.
If three points (h, 0), (a, b) and (0, k) lie on a line, show that \(\frac{a}{\boldsymbol{h}}+\frac{b}{\boldsymbol{k}}\) = 1.
Answer.
If the points A (h, 0), B (a, b) and C (0, k) lie on a line, then
Slope of AB = Slope of BC
\(\frac{b-0}{a-h}=\frac{k-b}{0-a}\)

⇒ \(\frac{b}{a-h}=\frac{k-b}{-a}\)

⇒ – ab = (k – b) (a – h)
⇒ – ab = ka – kh – ab + bh
⇒ ka + bh = kh
On dividing both sides by kh, we obtain
\(\frac{k a}{k h}+\frac{b h}{k h}=\frac{k h}{k h}\)

\(\frac{a}{h}+\frac{b}{k}\) = 1

Hence \(\frac{a}{h}+\frac{b}{k}\) = 1.

Question 14.
Consider the given population and year graph. Find the slope of the line AB and uaing it, find what will be the peculation in the year 2010?

Answer.
Since line AB passes through points A(1985, 92) and B(1995, 97), its slope is = \(\frac{97-92}{1995-1985}=\frac{5}{10}=\frac{1}{2}\)

Let y be the population in the year 2010.
Then, according to the given graph, line AB must pass through point C (2010, y)
∴ Slope of AB = Slope of BC

⇒ \(\frac{1}{2}=\frac{y-97}{2010-1995}\)

⇒ \(\frac{1}{2}=\frac{y-97}{15}\)

⇒ \(\frac{15}{2}\) = y – 97
⇒ y – 97 = 7.5
⇒ y = 7.5 + 97 = 104.5
Thus, the slope of line AB is \(\frac{1}{2}\), while in the year 2010, the population will be 104.5 crores.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 9 Sequences and Series Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 9 Sequences and Series Miscellaneous Exercise

Question 1.
Show that the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.
Answer.
Let a be the first term and d the common difference of an A.P.
Now, we want to prove that
T_{m + n} + T_{m – n} = 2T_m
L.H.S.= T_{m + n} + T_{m – n}
= [a + (m + n – 1) d] + [a + (m – n – 1)d]
= 2a + (m + n -1 + m – n – 1) d
= 2a + (2m – 2) d
= 2 (a + (m – 1) d]
= 2T_m R.H.S.
Hence proved.

Question 2.
If the sum of three numbers in AP is 24 and their product is 440, find the numbers.
Answer.
Let the three numbers in A.P. be a – d, a, and a + d.
According to the given information,
(a – d) + (a) + (a + d) = 24
3a = 24
⇒ a = 8
and (a – d) a (a + d) = 440 ………………(ii)
(8 – d) (8) (8 + d) = 440
(8 – d) (8 + d) = 55
= 64 – d² = 55
d² = 64 – 55 = 9
d = ±3
Therefore, when d = 3, the numbers are 5, 8 and 11 and when d = – 3, the numbers are 11, 8 and 5.
Thus, the three numbers are 5, 8 and 11.

Question 3.
Let the sum of n, 2n, 3n terms of an A.P. be S₁, S₂ and S₃, respectively, show that S₃ = 3 (S₂ – S₁).
Answer.
Let a be the first term and d common difference of an A.P. Therefore
S₁ = \(\frac{n}{2}\) [2a + (n – 1) d] ……………(i)
S₂ = \(\frac{2 n}{2}\) [2a + (2n – 1) d] ………………(ii)
S₃ = \(\frac{3 n}{2}\) [2a + (3n – 1) d] ……………….(iii)
Now from eq.s (i) and (ii) we have
S₂ – S₁ = \(\frac{2 n}{2}\) [2a + (2n – 1) d] – \(\frac{n}{2}\) [2a + (n – 1) d]
= \(\frac{n}{2}\) [4a + (4n – 2) d] – \(\frac{n}{2}\) [2a + (n – 1) d]

= \(\frac{n}{2}\) [4a + (4n – 2) d – 2a – (n – 1) d]

= \(\frac{n}{2}\) [2a + (3n – 1) d]

∴ 3 (S₂ – S₁) = \(\frac{3 n}{2}\) [2a + (3n – 1) d] = S₃

Hence, S₃ = 3 (S₂ – S₁)

Question 4.
Find the sum of all numbers between 200 and 400 which are divisible by 7.
Answer.
The number which are divisible by 7 between 200 and 400 are 203, 210, 217, ……………, 399.
Clearly, they form an A.P.
a = 203, d = 7 and T_n = 399
T_n = a + (n – 1) d
399 = 203 + (n – 1) 7
(n – 1) 7 = 399 – 203
(n – 1) 7 = 196
⇒ n – 1 = \(\frac{196}{7}\)
n – 1 = 28
⇒ n = 28 + 1
n = 29
S_n = \(\frac{n}{2}\) [2a + (n – 1) d]
S₂₉ = \(\frac{29}{2}\) [2 × 203 + (29 – 1) 7]
= \(\frac{29}{2}\) [406 + 28 × 7]
= \(\frac{29}{2}\) [406 + 196]
= \(\frac{29}{2}\) × 602 2 2 2
= 29 × 301 = 8729.

Question 5.
Find the sum of integers from 1 to 100 that are divisible by 2 or 5.
Answer.
The numbers from 1 to 100 which are divisible by 2 are 2, 4, 6, 8, …………, 100.
Clearly, they are in A.P., where a = 2 and d = 4 – 2 = 2
T_n = a + (n – 1) d
⇒ 100 = 2 + (n – 1)²
⇒ 100 – 2 = (n – 1)²
⇒ 98 = (n – 1)²
⇒ 49 = n – 1
n = 50
Therefore, sum of 50 numbers,
S₅₀ = \(\frac{50}{2}\) [2 × 2 + (50 – 1)²]
= 25 [4 + 49 × 2]
= 25 [4 + 98]
= 25 × 102
S₅₀ = 2250
Now, the numbers from 1 to loo which are divisible by 5 are 5, 10, 15, 20, 100.
Clearly, they are in A.P., where a = 5 and d = 10 – 5 = 5
T_n = a + (n – 1) d
100 = 5 + (n – 1) 5
⇒ 100 – 5 = (n – 1) 5
⇒ (n – 1) = \(\frac{95}{5}\)
n – 1 = 19
⇒ n = 19 + 1 = 20
Now, S_n = \(\frac{n}{2}\) [2a + (n – 1) d]
S₂₀ = \(\frac{20}{2}\) [2 × 5 + (20 – 1) 5]
= 10[10 + 19 × 5]
= 10 [10 + 95]
=10 (105)
S₂₀ = 1050 …(ii)
Now, the numbers from 1 to 100 which are divisible by 10 are 10, 20, 30, … 100.
Clearly, they are in A.P., where a = 10,
d = 20 – 10 = 10 and n = 10
S_n = [2a + (n – 1) d]
S₁₀ = \(\frac{10}{2}\) [2 × 10 + (10 – 1) 10]
= 5[20 + 9 × 10]
= 5[20 + 90]
= 5 × 110 = 550 …………….(iii)
Hence, required sum of integers from 1 to 100 which are divisible by 2
or 5 = 2550 + 1050 – 550 [using eqs. (i), (ii) and (iii)]
= 3600 – 550 = 3050.

Question 6.
Find the sum of all two digit numbers which when divided by 4, yields 1 as remninder.
Answer.
The sum of two digit numbers divisible by 4 yield 1 as remainder is 13 + 17 + 21 + ………… + 97.
Let the sum be denoted by S and let 97 be the nth term.
∴ T_n = a + (n – 1) d
97 = a + (n – 1) d
= 13 + (n – 1) 4
⇒ 97 = 13 + 4n – 4
⇒ 97 – 9 = 4n
⇒ n = 22
∴ The sum, S_n = 13 + 17 + 21 + ………….+ 97
∴ S_n = \(\frac{n}{2}\) [2a + (n – 1)d]
= \(\frac{22}{7}\) [2 × 13 + (22 – 1) × 4]
= 11 [26 + 21 × 4]
= 11 [26 + 84]
= 11 × 110 = 1210

Question 7.
If f is a function satisfying f(x + y) = f(x) f(y) for all x, y ∈ N such that f(1) = 3 and f(x) = 120, fInd the value of n.
Answer.
It is given that,
f(x + y) = f(x) × f(y) for all x, y ∈ N
f(1) = 3
Taking x = y = 1 in eq. (i), we obtain
f(1 + 1) = f(2) = f(1)
f(1) = 3 × 3 = 9
Similarly, f(1 + 1 + 1) = f(3) = f(1 + 2) = f(1) f(2) = 3 × 9 = 27
f(4) = f(1 + 3) = f(1) f(3) = 3 × 27 = 81
f(1), f(2), f(3) , that is 3, 9, 27, …………, forms a G.P. with both the first term and common ratio equal to 3.
It is known that, S_n = \(\frac{a\left(r^{n}-1\right)}{r-1}\)
It is given that, \(\sum_{x=1}^{n}\) f(x) = 120
∴ 120 = \(\frac{3\left(3^{n}-1\right)}{3-1}\)
⇒ 120 = \(\frac{3}{2}\) (3ⁿ – 1)
⇒ 3ⁿ – 1 = 80
⇒ 3ⁿ = 81 = 3
∴ n = 4
Thus, the value of n is 4.

Question 8.
The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.
Answer.
Let the sum of n terms of the G.P. be 315.
It is known that, Sⁿ = \(\frac{a\left(r^{n}-1\right)}{r-1}\)
It is given that the first term a is 5 and common ratio r is 2.
315 = \(\frac{5\left(2^{n}-1\right)}{2-1}\)
⇒ 2n = 64 = (2)⁶
⇒ n = 6
∴ Last term of the G.P.= 6th term
= ar^{6 – 1} = (5) (2)⁵ = (5) (32) = 160
Thus, the last term of the G.P. is 160.

Question 9.
The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.
Answer.
Let a and r be the first term and the common ratio of the G.P. respectively.
a = 1
a₃ = ar² = r²
a₅ = ar⁴ = r⁴
According to the question,
r² + r⁴ = 90
r⁴ + r² – 90 = 0
r² = \(\frac{-1 \pm \sqrt{1+360}}{2}\)

= \(\frac{-1 \pm \sqrt{361}}{2}=\frac{-1 \pm 19}{2}\)

= – 10 or 9

∴ r = ± 3 (Taking real roots)
Thus, the common ratio of the G.P. is ± 3.

Question 10.
The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.
Answer.
Let the three numbers in G.P. be a, ar and ar².
From the given condition, a + ar + ar² = 56
a (1 + r + r²) = 56
a = \(\frac{56}{1+r+r^{2}}\) …………………(i)
a – 1, ar – 7, ar² – 21 forms an A.P.
∴ (ar – 7) – (a – 1) = (ar² – 21) – (ar – 7)
⇒ ar – a – 6 = ar² – ar – 14
⇒ ar² – 2ar + a = 8
⇒ ar² – ar – ar + a = 8
a(r² + 1 – 2r) = 8
a(r – 1)² = 8 ………….(ii)
\(\frac{56}{1+r+r^{2}}\) (r – 1)² = 8 [using eq. (1)]
⇒ 7 (r² – 2r + 1) = 1 + r + r²
⇒ 7r² – 14r + 7 – 1 – r – r² = 0
⇒ 6r² – 15r + 6 = 0
⇒ r² – 12r – 3r + 6 = 0
⇒ 6r (r – 2) – 3 (r – 2) = 0
⇒ (6r – 3)(r – 2) = 0
∴ r = 2, \(\frac{1}{2}\)
When r = 2, a = 8; When r = \(\frac{1}{2}\), a = 32
Therefore, when r = 2, the three numbers in G.P. are 8, 16 and 32.
When r = \(\frac{1}{2}\), the three numbers in G.P. are 32, 16 and 8.
Thus, in either case, the three required numbers are 8, 16, and 32.

Question 11.
A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.
Answer.
Let the G.P. be T₁, T₂, T₃, T₄, …………….T_2n
Number of terms = 2n
According to the given condition,
T₁ + T₂ + T₃ + ………….. + T_2n = 5 [T₁ + T₃ + ……………. + T_{2n – 1}]
⇒ T₁ + T₂ + T₃ + ………….. + T_2n – 5 [T₁ + T₃ + …………… + T_{2n – 1}] = o
⇒ T₂ + T₄ + ………….. + T_2n = 4 [T₁ + T₃ + ………… + T_{2n – 1}]
Let the G.P. be a, ar, ar², ar³ ………..
∴ \(\frac{\ {ar}\left(r^{n}-1\right)}{r-1}\) = \(\frac{4 \times a\left(r^{n}-1\right)}{r-1}\)
⇒ ar = 4a
r = 4
Thus, the common ratio of the G.P. is 4.

Question 12.
The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.
Answer.
Let the A.P. be a, a + d, a + 2d, a + 3d,… a + (n – 2) d, a + (n – 1 )d
Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6d
Sum of last four terms = [a + (n – 4) d] + [a +(n – 3) d] + [a + (n – 2) d] + [(a + n – 1) d]
= 4a + (4n – 10)d
According to the given condition,
4a + 6d = 56
⇒ 4 (11) + 6d = 56 [Since a = 11 (given)]
⇒ 6d = 12
⇒ d = 2
4a + (4n – 10) d = 112
⇒ 4(11) + (4n – 10) 2 = 112
⇒ (4n – 10)2 = 68
⇒ 4n – 10 = 34
4n = 44
n = 11
Thus, the number of terms of the A.P. is 11.

Question 13.
If \(\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}=\frac{c+d x}{c-d x}\), then show that a, b, c and d are in G.P.
Answer.
It is given that \(\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}\)
⇒ (a + bx) (b – cx) = (b + cx) (a – bx)
⇒ ab – acx + b² x – bcx² = ab – b²x + acx – bcx²
⇒ 2b²x = 2acx
⇒ b² = ac
\(\frac{b}{a}=\frac{c}{b}\) ……………….(i)

Also, \(\frac{b+c x}{b-c x}=\frac{c+d x}{c-d x}\)
⇒ (b + cx) (c – dx) = (b – cx) (c + dx)
⇒ bc – bdx + c²x – cdx² = bc + bdx – c²x – cdx²
⇒ 2 c²x = 2 bdx
⇒ c² = bd
⇒ \(\frac{c}{d}=\frac{d}{c}\) ………….. (ii)
From eqs. (i) and (ii), we obtain \(\frac{b}{a}=\frac{c}{b}=\frac{d}{c}\)
Thus, a, b, c and d are in G.P.

Question 14.
Let S be the sum, P the product and R the sum of reciprocal of n terms in a G.P. Prove that P² Rⁿ = Sⁿ.
Answer.
Let the G.P. be a, ar, ar², ar³, ………….., ar^{n – 1}
According to the given information,
S = \(\frac{a\left(r^{n}-1\right)}{r-1}\)
P = aⁿ r^{\(\frac{n(n-1)}{2}\)}
[∵ Sum of n natural numbers is n \(\frac{(n+1)}{2}\)]

Hence P² Rⁿ = Sⁿ.

Question 15.
The th, qth and rth terms of an AP. are a, b, c respectively. Show that(q – r)a + (r – p) b + (p – q) c = 0.
Answer.
Let A be the first term and d be the common difference.
Since, T_p = a
⇒ A + (p – 1) d = a ……………(i)
T_q = b
⇒ A + (q – 1) d = b …………….(ii)
and T_r = c
⇒ A + (r – 1) d = c ……………..(iii)
(i) On multiplying eq. (i) by (q – r), eq. (ii) by (r – p) and eq. (iii) by (p – q),we get
(q – r) A + (p – 1) (q – r) d = a (q – r) …………..(iv)
(r – p) A + (q – 1) (r – p) d = b (r – p) …………….(v)
and (p – q) A + (r – 1) (p – q) d = c (p – q) ……………(vi)
On adding eqs. (iv), (v) and eq. (vi), we get
(q – r) A + (p – 1) (q – r) d + (r – p) A + (q – 1) (r – p) d + (p – q) A + (r – 1) (p – q) d = a (q – r) + b (r – p) + c (p – q)
⇒ A [(q – r) + (r – p) + (p – q)] + (p – 1) (q – r) + (q – 1) (r – p) + (r – 1) (p – q)] d = a (q – r) + b (r – p) + c (p – q)
A(0) + (0)d = a (q – r) + b (r – p) + c (p – q)
a (q – r) + b (r – p) + c (p – q) = 0
Hence proved.

Question 16.
If \(a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right)\) are in A.P.,prove that a, b, c are in A.P.
Answer.
It is given that \(a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right)\) are in A.P.

∴ \(b\left(\frac{1}{c}+\frac{1}{a}\right)-a\left(\frac{1}{b}+\frac{1}{c}\right)=c\left(\frac{1}{a}+\frac{1}{b}\right)-b\left(\frac{1}{c}+\frac{1}{a}\right)\)

⇒ \(\frac{b(a+c)}{a c}-\frac{a(b+c)}{b c}=\frac{c(a+b)}{a b}-\frac{b(a+c)}{a c}\)

⇒ \(\frac{b^{2} a+b^{2} c-a^{2} b-a^{2} c}{a b c}=\frac{c^{2} a+c^{2} b-b^{2} a-b^{2} c}{a b c}\)

⇒ b²a – a²b + b²c – a²c = c²a – b²a + c²b – b²c
⇒ ab (b – a) + c (b² – a²) = a (c² – b²) + bc (c – b)
⇒ ab (b – a) + c (b – a) (b + a) = a (c – b) (c + b) + bc (c – b)
⇒ (b – a) (ab + cb + ca) = (c – b) (ac + ab + bc)
⇒ b – a = c – b
Thus, a, b and c are in A.P.

Question 17.
If a, b, c, d are in G.P. prove that (aⁿ + bⁿ), (bⁿ + cⁿ), (cⁿ + dⁿ) are in G.P.
Answer.
It is given that a, b, c and d are in G.P.
∴ b² = ac ……………(i)
c² = bd …………….(ii)
ad = bc ……………..(iii)
It has to be proved that (aⁿ + bⁿ), (bⁿ + cⁿ), (cⁿ + dⁿ) are in G.P.
i.e., (bⁿ + cⁿ)² = (aⁿ + bⁿ) (cⁿ + dⁿ)
Consider L.H.S.
(bⁿ+ cⁿ)² = b²ⁿ + 2 bⁿcⁿ + c²ⁿ
= (b²)ⁿ + 2bⁿcⁿ + (c²)ⁿ
= (ac)ⁿ + 2bⁿcⁿ + (bd)ⁿ [using eqs. (i) and (ii)]
= aⁿcⁿ + bⁿcⁿ + bⁿcⁿ + bⁿdⁿ
= aⁿcⁿ + bⁿcⁿ + aⁿdⁿ + bⁿdⁿ [using eq. (iii)]
= cⁿ (aⁿ + bⁿ) + dⁿ (aⁿ + bⁿ)
= (aⁿ + bⁿ) (cⁿ + dⁿ) = R.H.S.
∴ (bⁿ + cⁿ)² = (aⁿ + bⁿ) (cⁿ + dⁿ)
Thus, (aⁿ + bⁿ), (bⁿ + cⁿ), and (cⁿ + dⁿ) are in G.P.

Question 18.
If a and b are the roots of x² – 3x + p = 0 and c, d are roots of x² – 12x + q = 0, where a, b, c, d form a G.P. Prove that (q + p) : (q – p) = 17 : 15.
Answer.
a, b are the roots of x² – 3x + p = 0
∴ a + b = 3, ab = p ……………. (i)
Again, c, d are the roots of xx² – 12x + q = 0
∴ c + d = 12 and cd = q
Also, a, b, c, d are in G.P.
Let r be its common ratio.
∴ b = ar, c = ar², d = ar³
Now a + b = a + ar = 3, [from eq. (i)]
and c + d = ar² + ar³ = 12, [from eq. (ii)]
Dividing, we get
\(\frac{a(1+r)}{a r^{2}(1+r)}=\frac{3}{12}=\frac{1}{4}\)

Question 19.
The ratio of the A.M. and G.M. of two positive numbers a and b, is m : n. Show that a : b (m + \(\sqrt{m^{2}-n^{2}}\)) : (m – \(\sqrt{m^{2}-n^{2}}\)).
Answer.

Question 20.
Ifa, b, c are in AP. a; b, c, d are in G.P. and \(\frac{1}{c}\), \(\frac{1}{d}\), \(\frac{1}{e}\) are in AP. prove that a, e, e are in GP.
Answer.
a, b, c are in A.P.
⇒ b = \(\frac{a+b}{2}\) ……………..(i)
b, c, d are in G.P.
⇒ c² = bd
\(\frac{1}{c}\), \(\frac{1}{d}\), \(\frac{1}{e}\) are in A.P.
⇒ \(\frac{2}{d}=\frac{1}{c}+\frac{1}{e}\)
d = \(\frac{2 c e}{c+e}\)
Putting the value of b and d from eq. (i) and (iii) in eq.(ii), we get
c² = \(\frac{a+c}{2} \cdot \frac{2 c e}{c+e}=\frac{c e(a+c)}{c+e}\)
c (c + e) = e (a + c)
c² + ce = ae + ce
c² = ae
∴ a, c, e are in G.P.

Question 21.
Find the sum of the following series up to n terms.
(i) 5 + 55 + 555 + ……………
(ii) .6 + .66 + .666 + ……………
Answer.
(i) We have, 5 + 55 + 555 + ……………… to n terms
= 5 (1 + 11 + 111 + ………… to n terms)
= \(\frac{5}{9}\) (9 + 99 +999 + ………… n terms)
[multiplying numerator and denominator by 9]
= \(\frac{5}{9}\) [(10 – 1) + (10² – 1) + (10³ – 1) + ………….. (10ⁿ – 1)]
= \(\frac{5}{9}\) [(10 + 10² + 10³ + …………… + 10ⁿ)] – (1 + 1 + …………. + 1) n terms]

= \(\frac{5}{9}\left[10\left(\frac{10^{n}-1}{10-1}\right)-n\right]\)

[∵ sum of G.P. = a \(\frac{\left(r^{n}-1\right)}{r-1}\)]

= \(\frac{5}{9}\left[\frac{10}{9}\left(10^{n}-1\right)-n\right]\)

= \(\frac{50}{81}\left(10^{n}-1\right)-\frac{5 n}{9}\)

(ii) We have 0.6 + 0.66 + 0.666 +… + to n terms
= 6 × 0.1 + 6 × 0.11 + 6 × 0.111 + ………………. + to n terms
= 6 [0.1 + 0.11 + 0.111 + ……………. + to n terms]
= \(\frac{6}{9}\) [0.9 + 0.99 + 0.999 + …………….. + to n terms]
[multiplying numerator and denominator by 9]

= \(\frac{2}{3}\left[\frac{9}{10}+\frac{99}{100}+\frac{999}{1000}+\ldots+\text { to } n \text { terms }\right]\)

= \(\frac{2}{3}\left[\left(1-\frac{1}{10}\right)+\left(1-\frac{1}{100}\right)+\left(1-\frac{1}{1000}\right)+\ldots \text { to } n \text { terms }\right]\)

= \(\frac{2}{3}\left[(1+1+1+\ldots n \text { terms })-\left(\frac{1}{10}+\frac{1}{10^{2}}+\frac{1}{10^{3}}+\ldots \text { to } n \text { terms }\right)\right.\)

= \(\frac{2}{3}\left[n-\frac{1}{10}\left\{\frac{1-\left(\frac{1}{10}\right)^{n}}{1-\frac{1}{10}}\right\}\right]\)

= \(\frac{2}{3}\left[n-\frac{1}{10}\left\{\frac{1-\left(\frac{1}{10}\right)^{n}}{\frac{9}{10}}\right\}\right]\)

[∵ sum of G.P. = a \(\frac{\left(1-r^{n}\right)}{(1-r)}\), |r| < 1]

= \(\frac{2}{3}\left[n-\frac{1}{9}\left\{1-\left(\frac{1}{10}\right)^{n}\right\}\right]\)

= \(\frac{2}{3}\) n – \(\frac{2}{27}\) (1 – 10^{– n}).

Question 22.
Find the 20th term of the series 2 × 4 + 4 × 6 + 6 × 8 + ……………..+ n terms.
Answer.
The given series is 2 × 4 + 4 × 6 + 6 × 8 + …………….. n terms
∴ nth term = a_n = 2n × (2n + 2) = 4n² + 4n
a20 = 4(20)² + 4(20)
= 4 (400) + 80 = 1600 + 80 = 1680
Thus, the 20th term of the series is 1680.

Question 23.
Find the sum of the first n terms of the series : 3 + 7 + 13 + 21 + 31 + …………….
Answer.
S_n = 3 + 7 + 13 + 21 + 31 + ………… + T_n …………….(i)
S_n = 3 + 7 + 13 + … + T_{n – 1} + T_n ……………(ii)
Then, eq. (i) – eq. (ii) given
0 = 3 + 4 + 6 + 8 + 10 + …………… to n terms T_n
T_n = 3 + (4 + 6 + 8 + 10 + …………….. to n – 1 terms )
3 + \(\frac{n-1}{2}\) [2 × 4 + (n – 1 – 1) 2]
= 3 + \(\frac{n-1}{2}\) [8 + 2n – 4]
= 3 + \(\frac{n-1}{2}\) [4 + 2n]
= 3 + (n – 1) (2 + n)
= 3 + n² + n – 2
= n² + n + 1
S_n = Σ (n² + n + 1)
= Σn² + Σn + 1 × n
= \(\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}\) + n
= \(\frac{n}{6}\) [2 n² + n + 2n + 1 + 3n + 3 + 6]
= \(\frac{n}{6}\) [2n² + 6n + 10]
= \(\frac{n}{6}\) (n² + 3n + 5).

Question 24.
If S₁, S₂, S₃ are the sum of first n natural numbers, their squares and their cubes, respectively, show that 9S₂² = S₃ (1 + 8S₁).
Answer.
S₁ is sum of the first n natural numbers.
∴ S₁ = Σn = \(\frac{n(n+1)}{2}\)

So is the sum of the cubes of first n natural numbers.
S₂ = Σn²
= \(\frac{n(n+1)(2 n+1)}{6}\)
S₃ is the sum of the cubes of first n natural numbers.

Question 25.
Find the sum of the following series up to n terms : \(\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+\ldots\)

Answer.

Question 26.
Show that \(\frac{1 \times 2^{2}+2 \times 3^{2}+\ldots+n \times(n+1)^{2}}{1^{2} \times 2+2^{2} \times 3+\ldots+n^{2} \times(n+1)}=\frac{3 n+5}{3 n+1}\).
Answer.
Let T_n and T_n be the nth terms of numerator and denominator respectively and S_n, S’_n be the respective sums of their n terms.
We have,
L.H.S. = \(\frac{1 \cdot 2^{2}+2 \cdot 3^{2}+\ldots+n(n+1)^{2}}{1^{2} \cdot 2+2^{2} \cdot 3+\ldots+n^{2}(n+1)}\)
nth term of numerator = n (n + 1)²
= n(n² + 2n + 1)
= n³ + 2 n² + n
T_n = n³ + 2 n² + n ………………(i)
and nth term of denominator = n² (n + 1)
= n³ + n² ………………(ii)

Hence Proved.

Question 27.
A farmer buys a used tractor for Rs. 12000. He pays Rs. 6000 cash and agrees to pay the balance in nnnual installments of Rs. 500 plus 12% interest on the unpaid amount. How much will be the tractor cost him?
Answer.
Total cost of the tractor = ₹ 12,000
Paid cost = ₹ 6,000
Balance = ₹ 6,000
No. of instalments @ ₹ 500 each = 12
Interest on first instalments = ₹ \(\left(\frac{6,000 \times 12 \times 1}{100}\right)\) = ₹ 720
First instalment = ₹ (500 + 720)
= ₹ 1220
Interest on second instalments = ₹ \(\left(\frac{5500 \times 12 \times 1}{100}\right)\) = ₹ 660
Second instalment = ₹ (500 + 660) = ₹ 1160
Third instalment = ₹ (500 + 600) = ₹ 1100 and so on
Total amount paid in instalments = ₹ (1220 + 1160 +1100 + ……………. to 12 terms)
Here, a = 1220,
d = 1160 – 1220 = – 60, n = 12
S = \(\frac{12}{2}\) [2 × 1220 + (12 – 1) (- 60)]
= 6 [2440 – 11 × 60]
= 6 [2440 – 660]
= 6 × 1780
= 10680 = ₹ 10680
Amount paid by farmer = ₹ (6000 +10680) = ₹ 16680.

Question 28.
Shamshad Ali buys a scooter for Rs. 22000. He pays Rs. 4000 cash and agrees to pay the balance in annual installment of Rs. 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?
Answer.
It is given that Shamshad Ali buys a scooter for Rs. 22000 and pays Rs. 4000 in cash.
∴ Unpaid amount = Rs. 22000 – Rs. 4000 = Rs. 18000
According to the given condition, the interest paid annually is 10% of 18000, 10% of 17000, 10% of 16000 … 10% of 1000
Thus, total interest to be paid
= 10% of 18000 + 10% of 17000 + 10% of 16000 + …………… + 10% of 1000
= 10% of (18000 + 17000 + 16000 + …………….. + 1000)
= 10% of (1000 + 2000 + 3000 + …………….. + 18000)
Here, 1000, 2000, 3000, ……………… 18000 forms an A.P. with first term and common difference both equal to 1000.
Let the number of terms be n.
∴ 18000 = 1000 + (n -1) (1000)
⇒ n = 18
∴ 1000 + 2000 + …. + 18000 = \(\frac{18}{2}\) [2(1000) + (18 – 1) (1000)]
= 9 [2000 + 17000] = 171000
∴ Total interest paid = 10% of (18000 + 17000 + 16000 + ………… + 1000)
= 10% of Rs. 171000
= Rs. 17100
∴ Cost of scooter = Rs. 22000 + Rs. 17100 = Rs. 39100.

Question 29.
A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when 8th set of letter is mailed.
Answer.
The numbers of letters mailed forms a G.P. : 4, 4², ………….., 4⁸
First term = 4,
Common ratio = 4,
Number of terms = 8
It is known that the sum of n terms of a GP. is given by S_n = \(\frac{a\left(r^{n}-1\right)}{r-1}\)
S₈ = \(\frac{4\left(4^{8}-1\right)}{4-1}\)

= \(\frac{4(65536-1)}{3}=\frac{4(65535)}{3}\)

= 4(21845) = 87380

It is given that the cost to mail one letter is 50 paisa.
∴ Cost of mailing 87380 letters = Rs. 87380 × \(\frac{50}{100}\) = Rs. 43690
Thus, the amount spent when 8th set of letter is mailed is Rs. 43690.

Question 30.
A man deposited Rs. 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount and also calculate the total amount after 20 years.
Answer.
It is given that the man deposited Rs. 10000 in a bank at the rate of 5% simple interest annually.
∴ Interest in first year = \(\frac{5}{100}\) × Rs. 10000 = Rs. 500
Amount in 15th year = Rs. 10000 + 500 + 500 + ………. + 500 (14 times)
= Rs. 10000 + 14 × Rs. 500
= Rs. 10000 + Rs. 7000
= Rs. 17000
Amount after 20 years = Rs. 10000 + 500 + 500 + …………… + 500 (20 times)
= Rs. 10000 + 20 × Rs. 500
= Rs. 10000 + Rs. 10000
= Rs. 20000

Question 31.
A manufacturer reckons that the value of a machine, which costs him Rs. 15625, will depreciate each year by 20%. Find the estimated value at the end of 5 years.
Answer.
Cost of machine = Rs. 15625
Machine depreciates by 20% every year.
Therefore, its value after every year is 80% of the original cost i.e., \(\frac{4}{5}\) of the original cost.
∴ Value at the end of 5 years = 15625 × \(\frac{4}{5}\) × \(\frac{4}{5}\) × ……… × \(\frac{4}{5}\) (5 times)
= 5 × 1024 = 5120
Thus, the value of the machine at the end of 5 years is Rs. 5120.

Question 32.
150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.
Answer.
Let x be the number of days in which 150 workers finish the work. According to the given information,
150x = 150 + 146 + 142 + …………… (x + 8) terms
The series 150 + 146 + 142 + ……………… (x + 8) terms is an A.P. with first term 150,
common difference = 4 and
number of terms as (x + 8).
⇒ 150x = \(\frac{x+8}{2}\) [2 (150) + (x + 8 – 1) (- 4)]
⇒ 150x = (x + 8) [150 + (x + 7) (- 2)]
⇒ 150x = (x + 8) (150 – 2x – 14)
⇒ 150x = (x + 8) (136 – 2x)
⇒ 75x = (x + 8) (68 – x)
⇒ 75x = 68x – x² + 544 – 8x
⇒ x² + 75x – 60x – 544 = 0
⇒ x² + 15x – 544 = 0
⇒ x² + 32x – 17x – 544 = 0
⇒ x(x + 32) – 17(x + 32) = 0
⇒ (x – 17) (x + 32) = 0
⇒ x = 17 or x = – 32
However, x cannot be negative.
∴ x = 17
Therefore, originally, the number of days in which the work was completed is 17.
Thus, required number of days = (17 + 8) = 25.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 9 Sequences and Series Ex 9.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.4

Question 1.
Find the sum to n terms of the series
1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + ……………
Answer.
Let S = 1 . 2 + 2 . 3 + 3 . 4 + 4 . 5 + …………….
Then, nth term,
T_n = n(n + 1) = n² + n
∴ T_n = n² + n
On taking summation from 1 to n on both sides we get

Question 2.
Find the sum to n terms of the series
1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …………..
Answer.
The given series is 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …………..
nth term a_n = n (n + 1) (n + 2)
= (n² + n) (n + 2) = n² + 3n² + 2n

Question 3.
Find the sum of n terms of the series 3 × 1² + 5 × 2² + 7 × 3² + ……………..
Answer.
The given series is 3 × 1² + 5 × 2² + 7 × 3² + ……………..
nth term a_n = (2n + 1) n²
= 2n³ + n²
∴ S_n = \(\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n}\left(2 k^{3}+k^{2}\right)=2 \sum_{k=1}^{n} k^{3}+\sum_{k=1}^{n} k^{2}\)

= \(2\left[\frac{n(n+1)}{2}\right]^{2}+\frac{n(n+1)(2 n+1)}{6}\)

Question 4.
Fmd the sum to n terms of the series \(\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}\) + …….
Answer.
Let the given series be
S = \(\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}\) + …….
Then, nth term T_n = \(\frac{1}{n(n+1)}\)
Now, we will split the denominator of the nth term into two parts or we will write T_n as the difference of two terms.

Question 5.
Find the sum to n terms of the series 5² + 6² + 7² + … + 20².
Answer.
The given series is 5² + 6² + 7² + … + 20²
nth term, a_n = (n + 4)²
= n² + 8n + 16

Question 6.
Find the sum to n terms of the series 3 × 8 + 6 × 11 + 9 × 14 + …………..
Answer.
The given series is 3 × 8 + 6 × 11 + 9 × 14 + ………….
a_n = (nth term of 3, 6, 9 ………..) × (nth term of 8, 11, 14, …………)
= (3n) (3n + 5)
= 9n² + 15n

Question 7.
Find the sum to n terms of the series 1² + (1² + 2²) + (1² + 2² + 3²) + ………….
Answer.
Let T_n denote the nth term, then
Tn = 1² + (1² + 2²) + (1² + 2² + 3²) + ………….

Question 8.
Find the sum to re terms of the series whose nth term is given by n (n + 1) (n + 4).
Answer.
a_n = n (n + 1) (n + 4)
= n(n² + 5n + 4)
= n³ + 5n² + 4n

Question 9.
Find the sum to re terms of the series whose nth term is given by tthe n² + 2n.
Answer.
a_n = n² + 2n
S_n = \(\sum_{k=1}^{n} k^{2}+2^{k}=\sum_{k=1}^{n} k^{2}+\sum_{k=1}^{n} 2^{k}\) …………..(i)
Consider \(\sum_{k=1}^{n} 2^{k}\) = 2¹ + 2² + 2³ + ……………

The above series 2, 2², 2³, … is a G.P. with both the first term and common ratio equal to 2.
\(\sum_{k=1}^{n} 2^{k}=\frac{(2)\left[(2)^{n}-1\right]}{2-1}\) = 2 (2ⁿ – 1) ……………(ii)
Therefore, from eqs. (i) and (ii), we obtain
S_n = \(\sum_{k=1}^{n} 2^{k}\) + 2(2ⁿ – 1)
= \(\frac{n(n+1)(2 n+1)}{6}\) + 2(2ⁿ – 1).

 

Question 10.
Find the sum to re terms of the series whose reth term is given by (2n – 1)².
Answer.
Given, nth term T_n = (2n – 1)²
⇒ T_n = 4 n² + 1 – 4n
Now, S = Σ T_n
= Σ (4n² + 1 – 4n)
= 4 Σn² + Σ 1 – 4 Σn

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 9 Sequences and Series Ex 9.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3

Question 1.
Find the 20th and nth terms of the G.P. \(\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \ldots\)
Answer.
The given G.P. is \(\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \ldots\).
Here, a = first term = \(\frac{5}{2}\)
r = Common ratio = \(\frac{\frac{5}{4}}{5}=\frac{1}{2}\)

a₂₀ = ar^{20 – 1}

= \(\frac{5}{2}\left(\frac{1}{2}\right)^{19}\)

= \(\frac{5}{(2)(2)^{19}}=\frac{5}{(2)^{20}}\)

a_n = ar^{n – 1}

= \(\frac{5}{2}\left(\frac{1}{2}\right)^{n-1}\)

= \(\frac{5}{(2)(2)^{n-1}}=\frac{5}{(2)^{n}}\)

Question 2.
Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.
Answer.
Common ratio, r = 2
Let a be the first term of the G.P.
∴ a₈ = ar^{8 – 1} = ar⁷
ar⁷ = 192
a(2)⁷ = 192
a(2)⁷ = (2)⁶ (3)
a = \(\frac{(2)^{6} \times 3}{(2)^{7}}=\frac{3}{2}\)
∴ a₁₂ = ar^{12 – 1}
= (\(\frac{3}{2}\)) (2)¹¹
= (3)(2)¹⁰ = 3072

Question 3.
The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q² =ps.
Answer.
Let a be the first term and r be the common ratio of the G.P.
According to the given condition.
a₅ = ar^{5 – 1}
= ar⁴ = p ………………(i)
a₈ = ar^{8 – 1}
= ar⁷ = q ……………(ii)
a₁₁ = ar^{11 – 1}
= ar¹⁰ = s ……………(iii)
Dividing equation (ii) by equation (i) we obtain
\(\frac{a r^{7}}{a r^{4}}=\frac{q}{p}\)

r³ = \(\frac{q}{p}\) ……………(iv)

Dividing equation (iii) by (ii), we obtain
\(\frac{a r^{10}}{a r^{7}}=\frac{s}{q}\)

r³ = \(\frac{s}{q}\) ……………….(v)
Equating the values of r³ obtained in eqs. (iv) and (v), we obtain
\(\frac{q}{p}=\frac{s}{q}\)
⇒ q² = ps
Thus, the given result is proved.

Question 4.
The 4th term of a G.P. is square of its second term, and the first term is – 3. Determine its 7th term.
Answer.
Let a be the first term and r be the common ratio of the G.P.
∴ a = – 3
It is known that,
a_n = a r^{n – 1}
∴ a₄ = ar³
= (- 3) r³
a₂ = ar¹
= (- 3) r
According to the given condition.
(- 3) r³ = [(- 3) r]²
⇒ – 3r³ = 9r²
⇒ r = – 3
a7 = ar^{7 – 1} = ar⁶
= (- 3) (- 3)⁶
= – (3)⁷ = – 2187
Thus, the seventh term of the G.P. is – 2187.

Question 5.
Which term of the following sequences :
(a) 2, 2√2, 4, ……….. is 128 ?
(b) √3, 3, 3√3, ………… is 729?
(c) \(\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots\) is \(\frac{1}{19683}\)?
Answer.
(a) The given sequence is 2, 2√2, 4, ………..
Here, a = 2 and r = \(\frac{2 \sqrt{2}}{2}\) = √2
Let the nth term of the given sequence be 128.
a_n = ar^{n – 1}
⇒ (2) (√2)^{n – 1} = 128
⇒ (2) \(\text { (2) } \frac{n-1}{2}\) = (2)⁷
⇒ \(\text { (2) } \frac{n-1}{2}+1\) = (2)⁷
∴ \(\frac{n – 1}{2}\) + 1 = 7
\(\frac{n – 1}{2}\) = 6
⇒ n – 1 = 12
⇒ n = 13
Thus, the 13th term of the given sequence is 128.

(b)

(c)

Question 6.
For what values of x, the numbers – \(\frac{2}{7}\), x, – \(\frac{7}{2}\) are in GP.?
Answer.
The given numbers are – \(\frac{2}{7}\), x, – \(\frac{7}{2}\)

⇒ x = √1
⇒ x = ± 1
Thus, fcr x = ±1 , the given numbers will be in GP.

Question 7.
Find the sum to 20 terms In the geometric progression 0.15, 0.015, 0.0015 ………
Answer.
The given G.P. is 0.15, 0.015, 0.0015, ………..
Here, a = 0.15 and r = \(\frac{0.015}{0.15}\) = 0.1
S_n = \(\frac{a\left(1-r^{n}\right)}{1-r}\)

S_n = \(\frac{0.15\left[1-(0.1)^{20}\right]}{1-0.1}\)

= \(\frac{0.15}{0.9}\left[1-(0.1)^{20}\right]\)

= \(\frac{1}{6}\) [1 – (0.1)²⁰].

Question 8.
Find the sum to n terms in the geometric progression √7, √21, 3√7 ……………
Answer.
The given G.P. is √7, √21, 3√7 ……………
Here, a = √7

Question 9.
Find the sum to n terms in the geometric progression 1, – a, a², – a³ ………… (if a ≠ 1)
Answer.
The given G.P. is 1, – a, a², – a³ …………
Here, first term = a₁ = 1
Common ratio = r = – a
S_n = \(\frac{a-{n}\left(1-r^{n}\right)}{1-r}\)

∴ S_n = \(\frac{1\left[1-(-a)^{n}\right]}{1-(-a)}=\frac{\left[1-(-a)^{n}\right]}{1+a}\)

Question 10.
Find the sum to n terms in the geometric progression x³, x⁵, x⁷ … (if x ≠ ± 1).
Answer.
The given G.P. is x³, x⁵, x⁷, ………….
Here, a = x³ and r = x²
S_n = \(\frac{a\left(1-r^{n}\right)}{1-r}\)

= \(\frac{x^{3}\left[1-\left(x^{2}\right)^{n}\right]}{1-x^{2}}=\frac{x^{3}\left(1-x^{2 n}\right)}{1-x^{2}}\).

Question 11.
Evaluate \(\sum-{k=1}^{11}\) (2 + 3^k)
Answer.

Question 12.
The sum of first three terms of a G.P. is – and their product is 1. Find the common ratio and the terms.
Answer.
Let the first three number of G.P. be \(\frac{a}{r}\), a and ar.
According to the question,
\(\frac{a}{r}\) + a + ar = \(\frac{39}{10}\) … (i)
and (\(\frac{a}{r}\)) × (a) × (ar) = 1
⇒ a³ = 1
⇒ a = 1
On putting the value of a = 1 in eq. (i), we get
\(\frac{1}{r}\) + 1 + r = \(\frac{39}{10}\)
\(\frac{1+r+r^{2}}{r}=\frac{39}{10}\)
⇒ 10 + 10r + 10r² = 39r
⇒ 10r² + 10r – 39r + 10 = 0
⇒ 10r² – 29r + 10 = 0
Now, factorising it by splitting the middle term, we get
10r² – 25r – 4r + 10 = 0
⇒ 5r (2r – 5) – 2 (2r – 5) = 0
⇒ (5r – 2)(2r – 5) = 0
⇒ 5r – 2 = 0 and 2r – 5 = 0
⇒ r = \(\frac{2}{5}\) and r = \(\frac{5}{2}\)

When a = 1 and r = \(\frac{2}{5}\), then numbers are
\(\frac{a}{r}=\frac{1}{\frac{2}{5}}=\frac{5}{2}\),
a = 1 and
ar = 1 × \(\frac{2}{5}\) = \(\frac{2}{5}\)
∴ \(\frac{5}{2}\), 1, \(\frac{2}{5}\).

When a = 1 and r = \(\frac{5}{2}\), then numbers are
\(\frac{a}{r}=\frac{1}{5}=\frac{2}{5}\);

a = 1 and ar = 1 × \(\frac{5}{2}\) = \(\frac{5}{2}\)
∴ \(\frac{2}{5}\), 1, \(\frac{5}{2}\).

Question 13.
How many terms of G.P. 3, 3², 3³, …………… are needed to give the sum 120?
Answer.
The given G.P. is 3, 3², 3³, …………
Let n terms of this G.P. be required to obtain the sum as 120.
S_n = \(\frac{a\left(r^{n}-1\right)}{r-1}\)
Here, a = 3 and r = 3
S_n = 120 = \(\frac{3\left(3^{n}-1\right)}{3-1}\)

⇒ 120 = \(\frac{3\left(3^{n}-1\right)}{3-1}\)
⇒ \(\frac{120 \times 2}{3}\) = 3ⁿ – 1
⇒ 3ⁿ – 1 = 80
⇒ 3ⁿ = 81
⇒ 3ⁿ = 3⁴
∴ n = 4
Thus, four terms of the given G.P. are required to obtain the sum as 120.

Question 14.
The sum of first three terms of a G.P. is 16 andthe sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.
Answer.
Let a be the first term and r be the common ratio, then

Question 15.
Given a G.P. with a = 729 and 7th term 64, determine S₇.
Answer.
a = 729, a₇ = 64
Let r be the common ratio of the G.P.
It is known that, a_n = ar^{n – 1}
a₇ = ar^{7 – 1} (729)r⁶

Question 16.
Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term.
Answer.
Let, first term = a and common ratio = r.
Given, sum of first two terms = – 4
⇒ a₁ + a₂ = – 4
⇒ a + ar = – 4
⇒ a (1 + r) = – 4
and fifth term = 4 × third term
a₅ = 4 × a₃
⇒ ar⁴ = 4ar²
r² = 4
r = ± 2
When r = 2, then from eq. (i), we get
a (1 + 2) = – 4
⇒ a = – \(\frac{4}{3}\)
Then, G.P. is – \(\frac{4}{3}\), – \(\frac{4}{3}\) × 2, – \(\frac{4}{3}\) × (2)², ……………
i.e., \(\frac{-4}{3}, \frac{-8}{3}, \frac{-16}{3}\)
When r = – 2, then from eq. (i), we get a
a (1 – 2)= – 4
⇒ – a = – 4
⇒ a = 4
Then, G.P. is 4, 4 × (- 2), 4 × (- 2)² … i.e, 4,- 8, 16, ………..

Question 17.
If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.
Answer.
Given, 4 th term,
T₄ = x
⇒ ar^{4 – 1} = x
⇒ ar³ = x …………………..(i)
10th term,
T₁₀ = y
⇒ ar^{10 – 1} = y
⇒ ar⁹ = y ………………….(ii)
and 16 th term, T₁₆ = z
⇒ ar^{16 – 1} = z
⇒ ar¹⁵ = z ………………(iii)
Now, multiplying eq. (j) by eq. (ill), we get
ar³ × ar¹⁵ = x × z
a² r^{3 + 15} = x × z
a² r¹⁸ = xz
(ar⁹)² = xz
∴ y² = xz [from eq. (ii)]
Therefore, x,y and z are in GP.

Question 18.
Find the sum to n terms of the sequence, 8, 88, 888, 8888 …………
Answer.
The given sequence is 8, 88, 888, 8888 ………………
This sequence is not a G.P.
However, it can be changed to G.P. by writing the terms as
S_n = 8 + 88 + 888 + 8888 + ……………. to n terms
= \(\frac{8}{9}\) [9 + 99 + 999 + 9999 + ………….to n terms]
= \(\frac{8}{9}\) [(10 – 1) + (10² – 1) + (10³ – 1) + (10⁴ – 1) + …………… to n terms]
= [(10 + 10² +………… n terrns) – (1 + 1 + 1 + ………….. n terms)]
= \(\frac{8}{9}\left[\frac{10\left(10^{n}-1\right)}{10-1}-n\right]=\frac{8}{9}\left[\frac{10\left(10^{n}-1\right)}{9}-n\right]\)
= \(\frac{80}{81}\) (10ⁿ – 1) – \(\frac{8}{9}\) n.

Question 19.
Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2,.
Answer.

Question 20.
Show that the products of the corresponding terms of the sequences a, ar, ar², … ar^{n – 1} and A, AR, AR², …….. AR^{n – 1} form a G.P. and find the common ratio.
Answer.
It has to be proved that the sequence, aA, arAR, ar²AR², ………. ar^{n – 1} AR^{n – 1}, forms a G.P.
\(\frac{\text { Second term }}{\text { First term }}=\frac{a r A R}{a A}\) = rR

\(\frac{\text { Third term }}{\text { Second term }}=\frac{a r^{2} A R^{2}}{a r A R}\) = rR
Thus, the above sequence forms a G.P. and the common ratio is rR.

Question 21.
Find four numbers forming a geometric progression in which third term is greater than the first term by 9, and the second term is greater than the 4th by 18.
Answer.
Let a be the first term and r the common ratio of G.P.
∴ nth term = T_n = ar^{n – 1}
⇒ T₂ = ar , T₃ = ar² and T₄ = ar³
Since third term is greater than the first by 9.
∴ T₃ = T₁ + 9
⇒ ar² = a + 9
Second term is greater than the 4th by 18.
T₂ = T₄ + 18
⇒ ar = ar³ + 18
⇒ ar³ = ar + 9r
From eqs. (ii) and (iii), we get
ar = ar + 9r + 18
⇒ 0 = 9r + 18
⇒ r = \(\frac{-18}{9}\) = – 2
Put = – 2 in (i), we get
a(- 2)² = a + 9
⇒ 4a = a + 9
⇒ 3a = 9
⇒ a = 3
T₂ = ar = 3 (- 2) = – 6
T₃ = ar²
= 3 (- 2)² = 12
T₄ = ar³
= 3 (- 2)³ = – 24
∴ Required terms are 3, – 6, 12 and – 24.

Question 22.
If the pth, qth and rth terms of a G.P. are a, b, and c, respectively. Prove that a^{q – r} b^{r – p} c^{p – q} = 1.
Answer.
Let A be the first term and R be the common ratio of the G.P.
According to the given information.
AR^{p – 1} = a
AR^{q – 1} = b
AR^{r – 1} = c
a^{q – r} b^{r – p} c^{p – q} = A^{q – r} x R^{(p – 1) (q – r)} x A^{r – p} x R^{(q – 1) (r – p)} x A^{p – q} x R^{(r – 1) (p – q)}
= A^{q – r + r – p + p – q} R^{(pq – pr – p + q) + (rq – r + p – pq) + (pr – p – qr + q)}
= A⁰ × R⁰ = 1
Thus, the given result is proved.

Question 23.
If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P² = (ab)ⁿ.
Answer.
The first term of the G.P. is a and the last term is b.
Therefore, the G.P. is a, ar, ar², ar³, …………. ar^{n – 1}, where r is the common ratio.
b = ar^{n – 1} …………..(i)
P = Product of n terms
= (a) (ar) (ar²) … (ar^{n – 1})
= (a × a × ……… a) (r × r² × ……….. r^{n – 1})
= aⁿ r^{1 + 2 + ………. + (n – 1)}
Here, 1, 2, ………. (n – 1) is an A.P.
∴

Thus, the given result is proved.

Question 24.
Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)th to (2n)th term is \(\frac{1}{r^{n}}\).
Answer.
Let a be the first term and r be the common ratio of the G.P.
Sum of first n terms = \(\frac{a\left(1-r^{n}\right)}{(1-r)}\)
Since there are n terms from (n + 1)th to (2n)th term,
sum of terms from (n + 1)th to (2n)th term
= \(\frac{a_{n+1}\left(1-r^{n}\right)}{(1-r)}\)
= \(\frac{a r^{n}\left(1-r^{n}\right)}{(1-r)}\) [∵ a_{n + 1} = ar^{n + 1 – 1} = arⁿ]

Thus, required ratio = \(\frac{a\left(1-r^{n}\right)}{(1-r)} \times \frac{(1-r)}{a r^{n}\left(1-r^{n}\right)}=\frac{1}{r^{n}}\)

Thus, the ratio of the sum of first n terms of G.P. to the sum of terms from (n + 1)th to (2n)th term is \(\frac{1}{r^{n}}\).

Question 25.
If a, b, c and d are in G.P. show that
(a² + b² + c²) (b² + c² + d²) = (ab + bc + cd)².
Answer.
Given, a, b, c, d are in G.P.
\(\frac{b}{a}=\frac{c}{b}=\frac{d}{c}\) = r (say)
⇒ b = ar, c = br, d = cr
⇒ b = ar, c = (ar)r, d = (br)r
⇒ b = ar, c = ar², d = br²
⇒ b = ar,c = ar²,d = (ar)r² = ar³
Now we have to prove that
(a² + b² + c²) (b² + c² + d²) = (ab + bc + cd)²
L.H.S. = (a² + b² + c²) (b² + c² + d²)
= (a² + a² r² + a² r⁴) (a² r² + a² r⁴ + a² r⁶)
= a² (1 + r² + r⁴) a² r² (1 + r² + r⁴)
= a⁴r² (1 + r² + r⁴)²
= [a² r (1 + r² + r⁴)]²
= [a² r + a²r³ + a² r⁵]²
=[a . ar + ar . ar² + ar² ar³]²
= [ab + be + cd]² [from eq. (i)]
= R.H.S.
Hence proved.

Question 26.
Insert two numbers between 3 and 81 so that the resulting sequence is GJ*.
Answer.
Let G₁ and G₂ be two numbers between 3 and 81 such that the Series, 3, G₁, G₂, 81, forms a G.P.
Let a be the first term and r be the common ratio of the G.P.
∴ 81 = (3)(r)³
r³ = 27
∴ r = 3
For r = 3
G₁ = ar = (3) (3) = 9
G₂ = ar² = (3) (3)² = 27

Question 27.
Find the value of n so that \(\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}\) may be the geometric mean between a and b.
Answer.

Hence proved.

Question 28.
The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio (3 + 2√2) : (3 – 2√2).
Answer.
Let the two numbers be a and b.
G.M. = √ab
According to the given condition,
a + b = √ab …………….(i)
(a + b)² = 36(ab)
Also, (a – b)² = (a + b)² – 4ab
= 36ab – 4ab = 32ab
a – b = √32 √ab
= 4 √2 √ab ……………….(ii)
Adding eqs. (i) and (ii), we obtain
26 = (6 + 4√2) √ab
⇒ a = (3 + 2√2)√ab
Substituting the value of a in (i), we obtain
b = 6√ab – (3 + 2√2) √ab
⇒ b = (3 – 2√2) √ab
\(\frac{a}{b}=\frac{(3+2 \sqrt{2}) \sqrt{a b}}{(3-2 \sqrt{2}) \sqrt{a b}}=\frac{3+2 \sqrt{2}}{3-2 \sqrt{2}}\)
Thus, the required ratio is (3 + 2√2) : (3 – 2√2).

Question 29.
If A and G be AM. and G.M., respectively between two positive numbers, prove that the numbers are A ± \(\sqrt{(A+G)(A-G)}\).
Answer.
let the numbers are α and β.
Given, sum of the roots,
\(\frac{\alpha+\beta}{2}\) = A [arithmetic mean]
α + β = 2A
and product of the root,
\(\sqrt{\alpha \beta}\) = G [geometric mean]
⇒ αβ = G²
Now, quadratic equation having roots α and β is
x² – (α + β)x + αβ = 0
⇒ x² – 2Ax + G² = 0

Hence proved.

Question 30.
The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th hour and nth hour?
Answer.
It is given that the number of bacteria doubles every hour.
Therefore, the number of bacteria after every hour will form a G.P.
Here, a = 30 and r = 2
∴ a₃ = ar²
= (30) (2)² = 120
Therefore, the number of bacteria at the end of 2nd hour will be 120.
a₅ = ar⁴
= (30) (2)⁴ = 480
The number of bacteria at the end of 4 th hour will be 480.
a_{n + 1} = arⁿ = (30) 2ⁿ
Thus, number of bacteria at the end of nth hour will be 30(2)ⁿ.

Question 31.
What will Rs. 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?
Answer.
The amount deposited in the bank is Rs. 500.
At the end of first year, amount = Rs. 500(1 + \(\frac{1}{10}\)) = Rs. 500 (1.1)
At the end of 2nd year, amount = Rs. 500 (1.1) (1.1)
At the end of 3rd year, amount = Rs. 500 (1.1) (1.1) (1.1) and so on .
Amount at the end of 10 years = Rs. 500 (1.1) (1.1) … (10 times)
= Rs. 500 (1.1)¹⁰.

Question 32.
If AM. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.
Answer.
Let the roots of the quadratic equation are α and β, then
(arithmetic mean) \(\frac{\alpha+\beta}{2}\) = 8 and
(geometric mean) \(\sqrt{\alpha \beta}\)= 5
⇒ α + β = 16 and αβ = 25
Now, if roots are a and p, then quadratic equation is x² – (Sum of roots) x + Product of roots = 0
⇒ x² – (α + β)x + αβ = 0
⇒ x² – 16x + 25 = 0.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 9 Sequences and Series Ex 9.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2

Question 1.
Find the sum of odd integers from 1 to 2001.
Answer.
The odd integers from 1 to 2001 are 1, 3, 5, … 1999, 2001.
This sequence forms an A.P.
Here, first term, a = 1
Common difference, d = 2
Here, a + (n – 1) d = 2001
⇒ 1 + (n -1) (2) = 2001
⇒ 2n – 2 = 2000
⇒ n = 1001
S_n = \(\frac{n}{2}\) [2a + (n – 1)d]
= \(\frac{1001}{2}\) [2 × 1 + (1001 – 1) × 2]
= \(\frac{1001}{2}\) [2 + 1000 x 2]
= \(\frac{1001}{2}\) × 2002
= 1001 × 1001
= 1002001
Thus, the sum of odd numbers from 1 to 2001 is 1002001.

Question 2.
Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.
Answer.
The natural numbers lying between 100 and 1000, which are multiples of 5 are 105, 110, …, 995.
Here, a = 105 and d = 5
a + (n – 1) d = 995
⇒ 105 + (n -1)5 = 995
⇒ (n – 1)5 = 995 – 105 = 890
⇒ n – 1 = 178
⇒ n = 179
S_n = \(\frac{179}{2}\) [2 (105) + (179 – 1) (5)]
= \(\frac{179}{2}\) [2 (105) + (178) (5)]
= 179 [105 + (89)5]
= (179) (105 + 445)
= (179) (550) = 98450
Thus, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5 is 98450.

Question 3.
In an A.P. the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is – 112.
Answer.
Let the first term of given AP be a and common difference be d.
We have, T₁ = a = 2
T₁ + T₂ + T₃ + T₄ + T₅ = [T₆ + T₇ + T₈ + T₉ + T₁₀]
Sum of 5 terms, where first term is a = \(\frac{1}{4}\) × sum of 5 terms, where first term is (a + 5d)
⇒ \(\frac{5}{2}\) [2a + (5 – 1) d] = \(\frac{1}{4}\) × \(\frac{5}{2}\) [2(a + 5d) + (5 – 1)d]
[∵ S_n = \(\frac{n}{2}\) [2a + (n – 1)d]]
\(\frac{5}{2}\) [2a + 4d] = \(\frac{1}{4}\) × \(\frac{5}{2}\) [2a + 10d + 4d]
\(\frac{5}{2}\) [2a + 4d] = \(\frac{1}{4}\) × \(\frac{5}{2}\) [2a + 14d]
⇒ 2a + 4d = \(\frac{1}{4}\) [2a + 14d]
2(2) + 4d = \(\frac{1}{4}\) [2 . (2) + 14d] [put a = 2]
4 + 4d = \(\frac{1}{4}\) [4 + 14d]
16 + 16d = 4 + 14d
16d – 14d = 4 – 16
2d = – 12
d = – 6
T₂₀ = a + (20 – 1) d
= 2 + 19 × (- 6)
= 2 – 114 = – 112
Hence proved.

Question 4.
How many terms of the A.P. – 6, – \(\frac{11}{2}\), – 5, … are needed to give the sum – 25?
Answer.
Let the sum of a terms of the given A.P. be -25.
It is known that, Sn = \(\frac{n}{2}\) [2a + (n – 1) d],
where n = number of terms, a = first term, and d common difference
Here, a = – 6
d = \(-\frac{11}{2}+6=\frac{-11+12}{2}=\frac{1}{2}\)
Therefore, we obtain
– 25 = \(\frac{n}{2}\left[2 \times(-6)+(n-1)\left(\frac{1}{2}\right)\right]\)
– 50 = n \(\left[-\frac{25}{2}+\frac{n}{2}\right]\)
⇒ – 100 = n (- 25 + n)
⇒ n² – 25n + 100 = 0
⇒ n² – 5n – 20n + 100 = 0
⇒ n (n – 5) – 20 (n – 5) = 0
⇒ n = 20 or 5.

Question 5.
In an A.P., if pth term is \(\frac{1}{q}\) and qth term is \(\frac{1}{p}\), prove that the sum of first pq terms is \(\frac{1}{2}\) (pq + 1), where pq ≠ q.
Answer.
It is known that the general term of an A.P. is
a_n = a + (n – 1) d
According to the given information,
pth term = a_p
= a + (p – 1)d = \(\frac{1}{q}\) …………..(i)
qth term = a_q
= a + (q – 1)d = \(\frac{1}{p}\) …………….(ii)
Subtracting eq. (ii) from eq. (i), we obtain
(p – 1) d – (q – 1) d = \(\frac{1}{q}-\frac{1}{p}\)
(p – 1 – q + 1) d = \(\frac{p-q}{p q}\)
⇒ (p – q) d = \(\frac{p-q}{p q}\)
d = \(\frac{1}{p q}\)
Putting the value of d in eq. (i), we obtain

Thus, the sum of first pq terms of the A.P. is \(\frac{1}{2}\) (pq + 1).

Question 6.
If the sum of a certain number, of terms of the A.P. 25, 22, 19, …is 116. Find the last term.
Answer.
Let the sum of n terms of the given A.P. be 116.
S_n = \(\frac{n}{2}\) [2a + (n – 1)d]
Here, a = 25 and d = 22 – 25 = – 3
S_n = \(\frac{n}{2}\) [2 × 25 + (n – 1) (- 3)]
⇒ 116 = \(\frac{n}{2}\) [50 – 3n + 3]
⇒ 232 = n(53 – 3n) = 53n – 3n²
3n² – 24n – 29n + 232 = 0
3n (n – 8) – 29 (n – 8) = 0
(n – 8) (3n – 29) = 0
n = 8 or n = \(\frac{29}{3}\)
However, n cannot be equal to \(\frac{29}{3}\).
Therefore, n = 8
∴ a₈ = Last term = a + (n -1) d = 25 + (8 – 1) (- 3)
= 25 + (7) (- 3) = 25 – 21 = 4
Thus, the last term of the A.P. is 4.

Question 7.
Find the sum of n terms of the A.P., whose kth term is 5k + 1.
Answer.
It is given that the kth term of the A.P. is 5k + 1
kth term = a_k = a + (k – 1) d
a + (k – 1) d = 5k +1
a + kd – d = 5k + 1
Comparing the coefficient of k, we obtain d = 5
a – d = 1
⇒ a – 5 = 1
⇒ a = 6
S_n = \(\frac{n}{2}\) [2a + (n – 1) d]
= \(\frac{n}{2}\) [2 (6) + (n – 1) (5)]
= \(\frac{n}{2}\) [12n + 5n – 5]
= \(\frac{n}{2}\) (5n + 7).

Question 8.
If the sum of n terms of an A.P. is (pn + qn 2), where p and q are constants, find the common difference.
Answer.
It is known that, S_n = \(\frac{n}{2}\) [2a + (n – 1)d]
According to the given condition,
\(\frac{n}{2}\) [2a + (n – 1 )d] = pn + qn²
\(\frac{n}{2}\) [2a + nd – d] = pn + qn²

na + n² \(\frac{d}{2}\) – n . \(\frac{d}{2}\) = pn + qn²
Comparing the coefficient of n² on both sides, we obtain
\(\frac{d}{2}\) = q
d = 2q
Thus, the common difference of the A.P. is 2q.

Question 9.
The sums of n terms of two arithmetic progressions are in the ratio 5n + 4 : 9n + 6. Find the ratio of their 18th terms.
Answer.
Let a₁, a₂ and d₁, d₂ be the first terms and the common difference of the first and second arithmetic progression respectively.
According to the given condition,
Sum of n terms of first A.P. 5n + 4
Sum of n terms of second A.P. 9n + 6

Question 10.
If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find sum of the first (p + q) terms.
Answer.
Let a and b be the first term and the common difference of the A.P. respectively.
Here, S_p = \(\frac{p}{2}\) [2a + (p – 1)d]
S_q = \(\frac{q}{2}\) [2a + (q – 1 )d]
According to the given condition,
\(\frac{p}{2}\) [2a + (p – 1) d] = \(\frac{q}{2}\) [2a + (q – 1) d]
p [2a + (p – 1) d] = q [2a + (n – 1) d]
2ap + pd (p – 1) d = 2aq + qd (q – 1) d
2a (p – q) + d[p(p – 1) – q (q – 1)] = 0
2a (p – q) + d[p² – p – q² + q] = 0
2a (p – q) + d [(p – q) (p + q) – (p – q)] = 0
2a (p – q) + d [(p – q) (p + q – 1)] = 0
2a + d (p + q – 1) = 0
d = \(\frac{-2 a}{p+q-1}\) ……………..(i)

∴ S_{p + q} = \(\frac{p+q}{2}\) [2a + (p + q – 1) d]
S_{p + q} = \(\frac{p+q}{2}\left[2 a+(p+q-1)\left(\frac{-2 a}{p+q-1}\right)\right]\) [from eq. (i)]
= \(\frac{p+q}{2}\) [2a – 2a] = 0
Thus, the sum of the first (p + q) terms of the A.P. is 0.

Question 11.
Sum of the first p, q and r terms of an A.P. are a, b and c, respectively.
Prove that \(\frac{a}{p}\) (q – r) + \(\frac{b}{q}\) (r – p) + \(\frac{c}{r}\) (p – q) = 0
Answer.
Given that, S_p = a, S_q = b, S_c = r
Let A be the first term and d be the common difference. Then,
S_p = \(\frac{p}{2}\) [2A + (p – 1) d] = a
2A + (p – 1) d = \(\frac{2 a}{p}\) ……………(i)

Question 12.
The ratio of the sums of m and n terms of an A.P. is m² : n². Show that the ratio of mth and nth term is (2m – 1): (2n – 1).
Answer.
Let the first term “be a and common difference be d.
Thus, S_m = \(\frac{m}{2}\) [2a + (m – 1)d] and S_n = \(\frac{n}{2}\) [2a + (n – 1)d]
According to the given condition.
\(\frac{S_{m}}{S_{n}}=\frac{m^{2}}{n^{2}}\)

\(\frac{\frac{m}{2}[2 a+(m-1) d]}{\frac{n}{2}[2 a+(n-1) d]}=\frac{m^{2}}{n^{2}} \Rightarrow \frac{2 a+(m-1) d}{2 a+(n-1) d}=\frac{m}{n}\)
2an + (mn – n) d = 2am + (mn – m)d
2an – 2am = (mn – m – mn + n)d
2a (n – m) = d (n – m)
⇒ d = 2a .
T_m = a + (m – 1) d = a + (m – 1) 2a
T_m = a + 2am – 2a
T_m = 2am – a
⇒ T_m = a (2m – 1) ……………..(i)
Also, Tn = a (2n – 1) ………………(ii)
On dividing eQuestion (i) by eQuestion (ii), we get
\(\frac{T_{m}}{T_{n}}=\frac{a(2 m-1)}{a(2 n-1)}=\frac{2 m-1}{2 n-1}\)
Hence proved.

Question 13.
If the sum of n terms of A.P. is 3n² + 5 n and its mth term is 164, find the value of m.
Answer.
Let a and b be the first term and the common difference of the A.P., respectively
a_m = a + (m – 1) d = 164 ……………(i)
Sum of n terms
Here, \(\frac{n}{2}\) [2a + nd – d] = 3n² + 5n
\(n a+\frac{n^{2} d}{2}-\frac{n d}{2}=3 n^{2}+5 n\)

\(\frac{n^{2} d}{2}+n\left(a-\frac{d}{2}\right)\) = 3n² + 5n

Comparing the coefficient of n² on both sides, we obtain
\(\frac{d}{2}\) = 3
⇒ d² = 2 × 3 = 6
Comparing the coefficient of n on both sides, we obtain
a – \(\frac{d}{2}\) = 5
a – \(\frac{6}{2}\) = 5
a = 5 + 3 = 8
Therefore, from eq. (i), we obtain
8 + (m – 1) 6 = 164
⇒ (m – 1) 6 = 164 – 8 = 156
⇒ m – 1 = 26
⇒ m = 27
Thus the value of m is 27.

Question 14.
Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.
Answer.
Let A₁, A₂, A₃, A₄ and A₅ be five numbers between 8 and 26 such that 8, A₁, A₂, A₃, A₄, A₅, 26 is an A.P.
Here, a = 8, b = 26, n = 7
Therefore, 26 = 8 + (7 – 1)d
6d = 26 – 8 = 18
d = 3
A₁ = a + d = 8 + 3 = 11
A₂ = a + 2d = 8 + 2 × 3 = 8 + 6 = 14
A₃ = a + 3d = 8 + 3 × 3 = 8 + 9 = 17
A₄ = a + 4d = 8 + 4 × 3 = 8 + 12 = 20
A₅ = a + 5d = 8 + 5 × 3 = 8 + 15 = 23.
Thus, the required five numbers between 8 and 26 are 11, 14, 17, 20 and 23.

Question 15.
If \(\frac{a^{n}+b^{n}}{a^{n-1}+b^{n-1}}\) is the A.M. etween a and b, then find the value of n.
Answer.
We know that, AM between a and b is \(\frac{a+b}{2}\).

On comparing the exponential powers, we get
n – 1 = 0
⇒ n = 1.

Question 16.
Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7th and (m – 1)th numbers is 5 : 9. Find the value of m.
Answer.
Let A₁, A₂, A₃, A₄, … A_m be m A.Ms.between 1 and 31.
Therefore, 1, A₁, A₂, A₃, ……… Am, 31 are in A.P.
Let d be the common difference of AP.
Here, the total number of terms is m + 2 and T_{m + 2} = 31
⇒ 1 + (m + 2 – 1) d = 31
⇒ (m + 1) d = 30
⇒ d = \(\frac{30}{m + 1}\) ……………….(i)
A₇ = T₈ = a + 7d

\(\frac{m+211}{31 m-29}=\frac{5}{9}\)
⇒ 9m + 1899 = 155m – 145
⇒ 146m = 2044
⇒ m = \(\frac{2044}{146}\) = 14
∴ m = 14

Question 17.
A man starts repaying a loan as first installment of Rs. 100. If he increases the installment by Rs. 5 every month, what amount he will pay in the 30th installment?
Answer.
The first installment of the loan is Rs. 100.
The second installment of the loan is Rs. 105 and so on.
The amount that the man repays every month forms an A.P.
The A.P. is 100, 105, 110, ………..
First term, a = 100 Common difference, d = 5
A₃₀ = a +(30 – 1)d
= 100 + (29) (5) = 100 + 145 = 245.
Thus, the amount to be paid in the 30th installment is Rs. 245.

Question 18.
The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°, find the number of the sides of ihe polygon.
Answer.
The angles of the polygon will form an A.P. with common difference d as 5° and first term a as 120°.
It is known that the sum of all angles of a polygon with n sides is 180° (n – 2).
∵ S_n = 180° (n – 2)
\(\frac{n}{2}\) [2a + (n -1) d] = 180° (n – 2)
\(\frac{n}{2}\) [240° + (n – 1) 5°] = 180 (n – 2)
n [240 + (n – 1) 5] = 360 (n – 2)
240n + 5n² – 5n = 360n – 720
⇒ 5n² + 235n – 360n + 720 = 0
⇒ 5n² – 125n + 720 = 0
⇒ n² – 25n + 144 = 0
⇒ n² – 16n – 9n + 144 = 0
⇒ n (n – 16) – 9 (n – 16) = 0
⇒ (n – 9) (n – 16) = 0
⇒ n = 9 or 16.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 16 Probability Ex 16.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 16 Probability Ex 16.3

Question 1.
Which of the following can not be valid assignment of probabilities for outcomes of sample space
S = {ω₁, ω₂, ω₃, ω₄, ω₅, ω₆, ω₇}

Answer.
(a)

Here, each of the numbers pω₁, is positive and less than 1.
Sum of probabilities = p (ω₁ ) + p (ω₂) + p (ω₃) + p(ω₄) + p (ω₅) + p (ω₆) + p (ω₇)
= 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1
Thus, the assignment is valid.

(b)

Here, each of the numbers pea f is positive and less than 1.
Sum of probabilities = p (ω₁ ) + p (ω₂) + p (ω₃) + p(ω₄) + p (ω₅) + p (ω₆) + p (ω₇)
= \(\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}=7 \times \frac{1}{7}\) = 1
Thus, the assignment is valid.

(c)

Sum of probabilities = p (ω₁) + p (ω₂) + p (ω₃) + p(ω₄) + p (ω₅) + p (ω₆) + p (ω₇)
= 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 = 2.8 ≠ 1
Thus, the assignment is not valid.

(d)

Here, p(ω₁) and p(ω₂) are negative.
Hence, the assignment is not valid.

(e)

Hence, the assignment is not valid.
p(ω₇) = \(\frac{15}{14}\) > 1.

Question 2.
A coin is tossed twice, what is the probability that atleast one tail occurs?
Answer.
When a coin is tossed twice, the sample space is given by S = (HH, HT, TH, TT}
Let A be the event of the occurrence of atleast one tail.
Accordingly, A – {HT, TH, TT}
∴ P(A) = \(\frac{\text Number of outcomes favourable}{\text { Total number of possible outcomes }}=\frac{n(A)}{n(S)}\)

= \(\frac{3}{4}\)

Question 3.
A die is thrown, find the probability of following events :
(i) A prime number will appear,
(ii) A number greater than or equal to 3 will appear,
(iii) A number less than or equal to one will appear,
(iv) A number more than 6 will appear,
(v) A number less than 6 will appear.
Answer.
The sample space of the given experiment is given by S = {1, 2, 3, 4, 5, 6}
(i) Let A be the event of the occurrence of a prime number. Accordingly, A = {2, 3, 5}
∴ P(A) = \(\frac{\text { Number of outcomes favourable to } A}{\text { Total number of possible outcomes }}=\frac{n(A)}{n(S)}\)

= \(\frac{3}{6}=\frac{1}{2}\)

Let B be the event of the occurrence of a number greater than or equal to 3. Accordingly, B = {3, 4, 5, 6}
∴ P(B) = \(\frac{\text { Number of outcomes favourable to } B}{\text { Total number of possible outcomes }}=\frac{n(A)}{n(S)}\)

= \(\frac{4}{6}=\frac{2}{3}\)

Let C be the event of the occurrence of a number les than or equal to one. Acordingly, C ={1}
∴ P(C) = \(\frac{\text { Number of outcomes favourable to } C}{\text { Total number of possible outcomes }}=\frac{n(C)}{n(S)}\)

= \(\frac{1}{6}\)

Let D be the event of the occurrence of a number greater than 6. Accordingly, D = {} = Φ
∴ P[D) = \(\frac{\text { Number of outcomes favourable to } D}{\text { Total number of possible outcomes }}=\frac{n(D)}{n(S)}\)

= \(\frac{0}{6}\) = 0

Let E be the event of the occurrence of a number less than 6. Accordingly, E = {1, 2, 3, 4, 5}
∴ P(E) = \(\frac{\text { Number of outcomes favourable to } E}{\text { Total number of possible outcomes }}=\frac{n(E)}{n(S)}\)

= \(\frac{5}{6}\).

Question 4.
A card is selected from a pack of 52 cards.
(a) How many points are there in the sample space?
(b) Calculate the probability that the card is an ace of spades.
(c) Calculate the probability that the card is
(i) an ace
(ii) black card.
Answer.
(a) when a card is selected from a pack of 52 cards, the number of possible outcomes is 52 i.e., the sample space contains 52 elements. Therefore there are 52 points in the sample space.

(b) Let A be the event in which the card drawn is an ace of spades. Accordingly, (n)A = 1
∴ P(A) = \(\frac{\text { Number of outcomes favourable to } A}{\text { Total number of possible outcomes }}=\frac{n(A)}{n(S)}\)

= \(\frac{1}{52}\)

(c) (i) Let E be the event in which the card drawn is an ace.
Since there are 4 aces in a pack of 52 cards,
Number of outcomes favourable to E n(E) = 4
∴ P(E) = \(\frac{\text { Number of outcomes favourable to } E}{\text { Total number of possible outcomes }}=\frac{n(E)}{n(S)}\)

= \(\frac{4}{52}=\frac{1}{13}\)

(ii) Let F be the event in which the card drawn is black.
Since there are 26 black cards in a pack of 52 cards, n(F) = 26

∴ P(F) = \(\frac{\text { Number of outcomes favourable to } F}{\text { Total number of possible outcomes }}=\frac{n(F)}{n(S)}\)

= \(\frac{26}{52}=\frac{1}{2}\).

Question 5.
A fair coin wIth 1 marked on one face and 6 on the other and a fair die are both tos,ed. Find the probability that the sum of numbers that turm up is
(i) 3
(ii) 12.
Answer.
Since the fair coin has 1 marked on one face and 6 on the other, and the die has six faces that are numbered 1,2, 3,4, 5, and 6, the sample space is given by
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Accordingly, n(S) = 12

(i) Let A be the event in which the sum of number that turn up is 3. Accordingly, A = {(1, 2)}
∴ P(A) = \(\frac{\text { Number of outcomes favourable to } A}{\text { Total number of possible outcomes }}=\frac{n(A)}{n(S)}\)

= \(\frac{1}{12}\)

(ii) Let B be the event in which the sum of numbers that turn up is 12. Accordingly, B = {(6, 6)}
∴ P(B) = \(\frac{\text { Number of outcomes favourable to } B}{\text { Total number of possible outcomes }}=\frac{n(B)}{n(S)}\)

= \(\frac{1}{12}\).

Question 6.
There are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman?
Answer.
There are four men and six women on the city council.
As one council member is to be selected for a committee at random, the sample space contains 10 (4 + 6) elements.
Let A be the event in which the selected council member is a woman.
Accordingly, n(A) = 6
∴ P(A) = \(\frac{\text { Number of outcomes favourable to } A}{\text { Total number of possible outcomes }}=\frac{n(A)}{n(S)}\)

= \(\frac{6}{10}=\frac{3}{5}\)

Question 7.
A fair coin is tossed four times, and a person win Re 1 for each head and lose Rs 1.50 for each tail that turns up. From the sample space calculate how many different amounts of money you can have after four tosses and the probability of have each of these amounts.
Answer.
There are five ways in which heads and tails appear.

(i) No head and 4 tail, appear
Money lost = Rs. 4 × 1.50 = Rs. 6.00
There is only 1 way when TTTT occurs.
No. of exhaustive cases = 2⁴ = 16
∴ Probaility of getting no head or 4 tails = \(\frac{1}{16}\).

(ii) When 1 head and 3 tail, appear
Money lost = Rs. (- 1 × 1 + 3 × 1.50) = Rs. 3.50
There are 4 ways when 1 head and 3 tails occur i.e., HTTT, THTT, TTHT, TTTH.
∴ Probaility of getting 1 head and 3 tails = \(\frac{4}{16}=\frac{1}{4}\).

(iii) When 2 head and 2 tail, appear
Money lost = Rs. (2 × 1.5 – 1 × 2) = Rs. (3 – 2) = Re. 1
2 heads and 2 tail may occur as HHTT, HTHT, HTTH, THHT, THTH, TTHH
Thus 2 head and 2 tails may appear in 6 ways
∴ Probability of getting 2 heads and 2 tails = \(\frac{6}{16}=\frac{3}{8}\)

(iv) When 3 head and 1 tail, appear
Money gained Rs. (3 × 1 – 1 × 1.5) = Rs. 1.50
3 heads and 1 tail may occurs as HHHT, HHTH, HTHH, THHH
∴ 3 heads and 1 tail appear in 4 ways
∴ Probability of getting 3 heads and 1 tail = \(\frac{4}{16}=\frac{1}{4}\).

(v) When all the heads appear
Money gained Rs. 4 × 1 = Rs. 4
4 heads occur as HHHH i.e., in one way.
Probability of getting 3 heads = \(\frac{1}{16}\).

Question 8.
Three coins are tossed once. Find the probability of getting
(i) 3 heads
(ii) 2 heads
(iii) at least 2 heads
(iv) at most 2 heads
(v) no head
(vi) 3 tails
(vii) exactly two tails
(viii) no tail
(ix) at most two tails.
Answer.
When three coins are tossed once, the sample space is given by
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
∴ Accordingly, n(S) = 8
It is known that the probability of an event A is given by,
P(A) = \(\frac{\text { Number of outcomes favourable to } A}{\text { Total number of possible outcomes }}=\frac{n(A)}{n(S)}\)

(i) Let B be the event of the occurrence of 3 heads. Accordingly, B={HHH}
∴ P(B) = \(\frac{n(B)}{n(S)}\)

= \(\frac{1}{8}\)

(ii) Let C be the event of the occurrence of 2 heads.
Accordingly, C = {HHT, HTH, THH}
∴ P(C) = \(\frac{n(C)}{n(S)}\)

= \(\frac{3}{8}\)

(iii) Let D be the event of the occurrence of at least 2 heads.
Accordingly, D = {HHH, HHT, HTH, THH}
∴ P(D) = \(\frac{n(D)}{n(S)}\)

= \(\frac{4}{8}=\frac{1}{2}\)

(iv) Let E be the event of the occurrence of almost 2 heads.
Accordingly, E = {HHT, HTH, THH, HTT, THT, TTH, TTT}
∴ P(E) = \(\frac{n(E)}{n(S)}\)

= \(\frac{7}{8}\)

(v) Let F be the event of the occurrence of no head.
Accordingly, F = {TTT}
∴ P(F) = \(\frac{n(F)}{n(S)}\)

= \(\frac{1}{8}\)

(vi) Let G be the event of the occurrence of 3 tails.
Accordingly, G = {TTT}
∴ P(G) = \(\frac{n(G)}{n(S)}\)

= \(\frac{1}{8}\)

(vii) Let H be the event of the occurrence of exacdy 2 tails.
Accordingly, H = {HTT, THT, TTH}
∴ P(H) = \(\frac{n(H)}{n(S)}\)

= \(\frac{3}{8}\)

(viii) Let I be the event of the occurrence of no tail.
Accordingly, I = {HHH}
∴ P(I) = \(\frac{n(I)}{n(S)}\)

= \(\frac{1}{8}\)

(ix) Let J be the event of the occurrence of atmost 2 tails.
Accordingly, J = {HHH, HHT, HTH, THH, HTT, THT, TTH}
∴ P(G) = \(\frac{n(J)}{n(S)}\)

= \(\frac{7}{8}\)

Question 9.
If \(\frac{2}{11}\) is the probability of an event, what is the probability of the event ‘not A’.
Answer.
It is given that P(A) = \(\frac{2}{11}\)
Accordingly, P (not A) = 1 – P(A)
= 1 – \(\frac{2}{11}\) = \(\frac{9}{11}\).

Question 10.
A letter is chosen at random from the word ‘ASSASSINATION’.
Find the probability that letter is (i) a vowel (ii) a consonant.
Answer. There are 13 letters in the word ASSASSINATION.
∴ Hence, n(S) = 13

(i) There are 6 vowels in the given word.
∴ Probability (vowel) = \(\frac{6}{13}\).

(ii) There are 7 consonants in the given word.
∴ Probability (consonant) = \(\frac{7}{13}\).

Question 11.
In a lottery, person choses six different natural numbers at random from 1 to 20, and if these six numbers match with the six numbers airedy fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game?
[Hint: order of the numbers is not important.]
Answer.
Total number of ways in which one can choose six different numbers from 1 to 20

= \({ }^{20} C_{6}=\frac{20 !}{6 !(20 !-6 !)}\)

= \(\frac{20 !}{6 ! 14 !}=\frac{20 \times 19 \times 18 \times 17 \times 16 \times 15}{1.2 .3 .4 .5 .6}\)

= 38760

Hence, there are 38760 combinations of 6 numbers.
Out of these combinations, one combination is already fixed by the lottery committee.
Required probability of winning the prize in the game = \(\frac{1}{38760}\).

Question 12.
Check whether the following probabilities P( A) and P(B) are consistently defined
(i) P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6
(ii) P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8 ;
Answer.
(i) Here, P(A ∩ B) > P(A)
So, given data is not consistent.

(ii) Here, P(A) = 05, P(B) = 0.4, P(A ∪ B) = 0.8
⇒ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ 0.8 = 0.5 + 0.4 – P(A ∩ B)
⇒ P(A ∩ B) = 0.9 – 0.8 = 0.1 < P(A) and P(B)
⇒ P(A) and P(B) are consistent.

Question 13.
Fill in the blanks in following table :

Answer.
(i) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{1}{3}+\frac{1}{5}-\frac{1}{15}\)

= \(\frac{5+3-1}{15}=\frac{7}{15}\)

(ii) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
0.6 = 0.35 + P(B) – 0.25
P(B) = 0.6 – 0.35 + 0.25 = 0.5

(iii) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
0.7 = 0.5 + 0.35 – P(A ∩ B)
P(A ∩ B) = 0.5 + 0.35 – 0.7 = 0.15

Question 14.
Given P(A) = \(\frac{3}{5}\) and P(B) = \(\frac{1}{5}\). Find P(A or B), if A and B are mutually exclusive events.
Answer.
Here, P(A) = \(\frac{3}{5}\), P(B) = \(\frac{1}{5}\)
For mutually exclusive events A and B,
P(A or B) = P(A) + P(B)
∴ P(A or B) = \(\frac{3}{5}\) + \(\frac{1}{5}\) = \(\frac{4}{5}\).

Question 15.
If E and F are events such that P(E) = \(\frac{1}{4}\), P(F) = \(\frac{1}{2}\) and P(E and F) = \(\frac{1}{8}\) find:
(i) P(E or F),
(ii) P (not E and not F).
Answer.
(i) P(E or F) = P(E ∪ F) = P(E) + P(F) – P(E ∩ F)
= \(\frac{1}{4}+\frac{1}{2}-\frac{1}{8}\)

= \(\frac{2+4-1}{8}=\frac{5}{8}\)

(ii) not E and not F = E ∩ F’ = (E ∪ F)’ (De Morgan’s Law)
P(not E and not F) = P(E ∪ F)’
= 1 – P(E ∪ F)
= 1 – \(\frac{5}{8}\) = \(\frac{3}{8}\).

Question 16.
Events E and F are such that P(not E or not F) = 0.25. State whether E and F are mutually exclusive.
Answer.
It is given that P (not E or not F) = 0.25 i.e., P(E ∪ F) = 0.25
⇒ P(E ∪ F)’ = 0.25
[∵ E ∪ F = (E ∩ F)’, De Morgan’s law]
Now, P(E ∩ F) = 1 – P(E ∩ F)’
⇒ P(E ∩ F) = 1 – 0.25
⇒ P(E ∩ F) = 0.75 ≠ 0
⇒ E ∩ F ≠ 0
Thus, E and F are not mutually exclusive.

Question 17.
A and Bare events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16. Determine
(i) P(not A),
(ii) P (not B)
(iii) P(A or B).
Answer.
It is given that P(A) = 0.42, P(B) = 0.48, P(A and B) = 0.16
(i) P(not A) = 1 – P(A) = 1 – 0.42 = 0.58
(ii) P(not B) = 1 – P(B) = 1 – 0.48 = 0.52
(iii) We know that P(A or B) = P(A) + P(B) – P(A and B)
P(A or B) = 0.42 + 0.48 – 0.16 = 0.74

Question 18.
In Class XI of a school 40% of the students study Mathematics and 30% study Biology. 10% of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology.
Answer.
Let A be the event in which the selected student studies Mathematics and B be the event in which the selected student studies Biology.
Accordingly, P(A) = 40% = \(\frac{40}{100}=\frac{2}{5}\);

P(B) = 30% = \(\frac{30}{100}=\frac{3}{10}\)

P(A and B) = 10% = \(\frac{10}{100}=\frac{1}{10}\)

We know that P(A or B) = P(A) + P(B) – P(A and B)

∴ P(A or B) = \(\frac{2}{5}+\frac{3}{10}-\frac{1}{10}=\frac{6}{10}\) = 0.6

Thus, the probability that the selected student will be studying mathematics or Biology is 0.6.

Question 19.
In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examination is 0.7. The probability of passing at least one of them is 0.95. What is the probability of passing both?
Answer.
Let A and B be the events of passing the first and second examinations respectively. .
P(A) = 0.8, P(B) = 0.7
Probability of passing at least one examination
= 1 – P(A’ ∩ B’) = 0.95 ……………….(i)
Now A’ ∩ B’ = (A ∪ B)’ (De Morgan’s law)
P(A’ ∩ B) = P(A ∪ B)’= 1 – (A ∪ B)
Putting the value in eQuestion (ii)
1 – [1 – P(A ∪ B)] = 0.95 or P(A ∪ B) = 0.95
Further P(A ∩ B) = P(A) + P(B) – P(A ∪ B)
= 0.8 + 0.7 – 0.95 = 1.5 – 0.95 = 0.55
Thus, probability that the student will pass in both the examinations = 0.55.

Question 20.
The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing the Hindi examination?
Answer.
Let A and B be the events of passing English and Hindi examinations respectively.
Accordingly, P(A and B) = 0.5, P(not A and not B) = 0.1 i.e.,
P(A’ ∩ B’) = 0.1
P(A)= 0.75
Now, (A ∪ B)’= (A’ ∩ BO [De Morgan’s law]
∴ P(A ∪ B)’ = P(A’ ∩ B’) = 0.1
P(A ∪ B) = 1 – P(A ∪ B)’= 1 – 0.1 = 0.9
We know that P(A or B) = P(A) + P(B) – P(A and B)
∴ 0.9 = 0.75 + P(B) – 0.5
⇒ P(B) = 0.9 – 0.75 + 0.5
⇒ P(B) = 0.65
Thus, the probability of passing the Hindi examination is 0.65.

Question 21.
In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. If one of these students is selected at random, find the probability that
(i) The student opted for NCC or NSS.
(ii) The student has opted neither NCC nor NSS.
(iii) The student has opted NSS but not NCC.
Answer.
Let A and B denote the students in NCC and NSS, respectively.
Given, n(A) = 30,
n(B) = 32
n(A ∩ B) = 24,
n(S) = 60 [∵ 24 students opted for both NCC and NSS i.e., they are common in both]

P(A) = \(\frac{30}{60}\) P(B) = \(\frac{32}{60}\) and P(A ∩ B) = \(\frac{24}{60}\)

Then, P(A ∩ B) = \(\frac{2}{12}\)

Now, required probability,
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{6}{12}+\frac{4}{12}-\frac{2}{12}\)

= \(\frac{8}{12}=\frac{2}{3}\)

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 16 Probability Ex 16.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 16 Probability Ex 16.2

Question 1.
A die is rolled. Let E be the event “die shows 4” and F be the event “die shows even number”. Are E and F mutually exclusive?
Answer.

If we roll a die, then all the possible outcomes will be
S= {1, 2, 3, 4, 5, 6}
E = Die shows 4 = {4}
F = Die shows an even number = (2, 4, 6}
⇒ E ∩ F = {4} ∩ {2, 4, 6} = {4}
⇒ E ∩ F ≠ Φ
Hence, E and F are not mutually exclusive.

Question 2.
A die is thrown. Describe the following events:
(i) A: a number less than 7
(ii) B: a number greater than 7
(iii) C: a multiple of 3
(iv) D: a number less than 4
(v) E: an even number greater than 4
(vi) F: a number not less than 3.
Also find A ∪ B, A ∩ B, B ∪ C, E ∩ F, D ∩ E, A – C, D – E, E ∩ F, F.
Answer.
When a die is throw, then sample space S = {1, 2, 3, 4, 5, 6}
(i) A : a number less than 7 = {1, 2, 3, 4, 5, 6}
(ii) B: a number greater than 7 = {} = {Φ}
(iii) C : a multiple of 3 = {3, 6}
(iv) D: a number less than 4 = {1, 2, 3}
(v) E : an even number greater than 4 = {6}
(vi) F : a number not less than 3 = {3, 4, 5, 6}
Now, A ∪ B = The elements which are in both A and B
{1, 2, 3, 4, 5, 6} ∪ – Φ = {1, 2, 3, 4, 5, 6}

A ∩ B – The elements which are common in both A and B
= {1, 2, 3, 4, 5, 6} ∩ Φ = Φ

B ∪ C = The elements which are in both B and C
= { } ∪ {3, 6} = {3, 6}

E ∩ F = The elements which are common in both E and F
= {6} ∩ {3, 4, 5, 6} = {6}

D ∩ E = The elements which are common in both D and E
= {1, 2, 3,} ∩ {6} = Φ

A – C = The elements which are in A but not in C
= {1, 2, 3, 4, 5, 6} – {3, 6} = {1, 2, 4, 5}

D – C = The elements which in D but not in E
= {1, 2, 3} – {6} = {1, 2, 3}

F’ = (S – F) = {1, 2, 3, 4, 5, 6} – {3, 4, 5, 6} = {1, 2}
and E ∩ F’ = E ∩ (S – F)
= {6} ∩ {1, 2} = Φ.

Question 3.
An experiment involves rolling a pair of dice and recording the numbers that come up. Describe the following events:
A : the sum is greater than 8,
B : 2 occurs on either die
C : The sum is at least 7 and a multiple of 3.
Which pairs of these events are mutually exclusive?
Answer.
When a pair of dice is rolled, the sample space is given by S = {(x, y) : x, y = 1, 2, 3, 4, 5, 6}

Accordingly,
A = {(3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}
B = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (1, 2), (3, 2), (4, 2), (5, 2), (6, 2)}
C = {(3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}
It is observed that
A ∩ B = Φ
B ∩ C = Φ
C ∩ A = {(3, 6), (4, 5), (5, 4), (6, 3), (6, 6)}
Hence, events A and B and events B and C are mutually exclusive.

Question 4.
Three coins are tossed once. Let A denote the event “three heads show”, B denote the event “two heads and one tail show”, C denote the event “three tails show” and D denote the event “a head shows on the first coin”.
Which events are
(i) mutually exclusive?
(ii) simple?
(iii) compound?
Answer.
When three coins are tossed, the sample space is given by S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TIT}
Accordingly,
A = {HHH},
B = {HHT, HTH, HTT},
C = (TIT),
D = {HHH, HHT, HTH, HTT}
We now observe that
A ∩ B = Φ A ∩ C = Φ; B ∩ C = Φ; C ∩ D = Φ
A ∩ B ∩ C = Φ

(i) Event A and B; event A and C; event B and C; event C and D and event A, B,C are all mutually exclusive.
(ii) If an event has only one sample point of a sample space, it is called a simple event, Thus, A and C are simple events.
(iii) If an event has more than one sample point of a sample space, it is called a compound event. Thus, B and D are compound events.

Question 5.
Three coins are tossed. Describe
(i) Two events which are mutually exclusive.
(ii) Three events which are mutually exclusive and exhaustive.
(iii) Two events, which are not mutually exclusive. 1
(iv) Two events which are mutually exclusive but not exhaustive,
(v) Three events which are mutually exclusive but not exhaustive.
Answer.
When three coins are tossed, then the sample space S is S={HHH, HHT, HTH, HTT, THH, THT, TTH, TIT}
(i) Two events A and B which are mutually exclusive are
A : “getting at least two heads” and
B : “getting at least two tails”.

(ii) Three events A, B and C which are mutually exclusive and exhaustive are
A : “getting at most one head”
B : “getting exactly two heads”
C : “getting exactly three heads”.
Alternatively A : “getting no head”
B : “getting exactly on head”
C : “getting at least two heads”.

(iii) Two events A and B which are not mutually exclusive are
A : “getting at most two tails” and
B : “getting exactly two heads” or “getting exacdy two tails”.

(iv) Two events A and B which are mutually exclusive but not exhaustive are
A : “getting exactly one head” and
B : “getting exactly two heads”.

(v) Three events A, B and C which are mutually exclusive but not exhaustive are
A : “getting exacdy one tail”
B : “getting exacdy two tails” and
C : “getting exacdy three tails”.

Question 6.
Two dice are thrown. The events A, B and C are as follows:
A: getting an even number on the first die.
B: getting an odd number on the first die.
C: getting the sum of the numbers on the dice ≤ 5
Describe the events
Answer.
When two dice are thrown, the sample space is given by
S = {(x, y): x, y 1, 2, 3, 4, 5, 6}
= [(1, 1), (1, 2), (1, 3),(1, 4), (1, 5) ,(1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)]

Accordingly,
A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3) (4, 4), (4, 5), (4, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}.

B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}.

C = {(1 ,1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}

(i) A’ = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)} = B

(ii) Not B = B’ = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
= A.

(iii) A or B = A ∪ B
= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2),(4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} = S

(iv) A and B = A ∩ B = Φ

(v) A but not C = A – C
= {(2, 4), (2, 5), (2, 6), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6,6)}

(vi) B or C = B ∪ C
= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}

(vii) B and C = B ∩ C = {(1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2)}

= {(1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

∴ A ∩ B’ ∩ C’ = A ∩ A ∩ C’ = A ∩ C’
= {(2, 4), (2, 5), (2, 6), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}.

Question 7.
Refer to question 6 above state true or false (given reason for your answer).
(i) A and B are mutually exclusive.
(ii) A and B are mutually exclusive and exhaustive.
(iii) A = B’
(iv) A and C are mutually exclusive.
(v) A and B’ are mutually exclusive.
(vi) A’, B’ and C are mutually exclusive and exhaustive.
Answer.
(i) True
∵ A = getting an even number on the first die
B = getting an odd number on the first die
⇒ A ∩ B = Φ
∴ A and B are mutually exclusive events.

(ii) True
A ∪ B = S, i.e., exhaustive. Also, A ∩ B = Φ

(iii) True
∴ B = getting an odd number on the first die
⇒ B’ = getting an even number on the first die = A
∴ A = B’

(iv) False
∵ A ∩ C = {(2, 1), (2, 2), (2, 3), (4, 1)} ≠ Φ.
So A and C are not mutually exclusive.

(v) False
∵ B’ = A
A ∩ B’ = A ∩ A = A ≠ Φ (∵ B’ = A)
So, A and B’ are not mutually exclusive.

(vi) False
A’ ∩ B’ = Φ
A’ ∩ B’ ∩ C = Φ and A’ ∪ B’ ∪ C = S
But A’ ∩ C = B ∩ C = {(1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2)} ≠ Φ
(∵ A’ = B)
and B’ ∩ C = A ∩ C = {(2, 1), (2, 2), (2, 3), (4, 1)} ≠ Φ (∵ B’ = A)
∴ A’, B’ and C are not mutually exclusive.