Class 10

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 6 Triangles Ex 6.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.1

Question 1.
Fill in the blanks using the correct word given in brackets:
(i) All circles are ………………. (congruent, similar).
Solution:
All circles are similar.

(ii) All squares are ………………. (similar, congruent).
Solution:
MI squares are similar.

(iii) All ………………. triangles are similar. (isosceles, equilateral).
Solution:
All equilateral triangles are similar.

(iv) Two polygons of the same number of sides are similar, if
(a) their corresponding angles are __________ and
Solution:
equal

(b) their corresponding sides are ………………. (equal, proportional).
Solution:
proportional.

Question 2.
Give two different examples of pair
(i) similar figures
(ii) non-similar figures.
Solution:
(i) 1. Pair of equilateral triangle are similar figures.
2. Pair of squares are similar figures.

(ii) 1. A triangle and quadrilateral form a pair of non-similar figures.
2. A square and rhombus form pair of non – similar figures.

Question 3.
State whether the following quadrilaterals are similar or not :-

Solution:
The two quadrilaterals in the figure are not similar because their corresponding angles are not equal.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 5 Arithmetic Progressions Ex 5.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4

Question 1.
Which term of the A.P. 121, 117, 113, …………. is its first negative term?
Solution:
Given A.P is 121, 117, 113, …
Here a = T₁ = 121 ;T₂ = 117; T₃ = 113
d = T₂ – T₁ = 117 – 121 = – 4
Using formula, T_n = a + (n – 1) d
T_n = 121 + (n – 1) (- 4)
= 121 – 4n + 4
= 125 – 4n.
According to question :—
T_n < 0
or 125 – 4n < 0
or 125 < 4n or 4n > 125.
or n > \(\frac{125}{4}\)
or n > 31\(\frac{1}{4}\).
But n must be integer, for first negative term.
∴ n = 32.
Hence, 32nd term be the first negative term of given A.P.

Question 2.
The sum of the third and the seventh term of an A.P. is 6 and their product ¡s 8. Find the sum of first sixteer
terms of an A.P.
Solution:
Let ‘a’ and ‘d’ be the first term and common diftèrence of given A.P.
According to 1st condition
T₃ + T₇ = 6
[a + (3 – 1)d] + [a + (7 – 1) d] = 6
∵ [T_n = a + (n – 1) d]
or a + 2d + a + 6d = 6
or 2a + 8d = 6
or a + 4d = 3 …………….(1)
According to 2nd condition
T₃ (T₇) = 8
[a + (3 – 1) d] [a + (7 – 1)d] = 8
∵ [T_n = a + (n – 1) d]
or (a + 2d) (a + 6d) = 8
or [3 – 4d + 2d] [3 – 4d + 6d] = 8
[Using (1), a = 3 – 4d]
or (3 – 2d) (3 + 2d) = 8
or 9 – 4d² = 8
or 4d² = 98
or d² = \(\frac{1}{4}\)
d = ± \(\frac{1}{2}\)

Case I:
When d = \(\frac{1}{2}\)
Putting d = \(\frac{1}{2}\) in (1), we get:
a + 4 (\(\frac{1}{2}\)) = 3
or a + 2 = 3
or a = 3 – 2 = 1
Using formula, S_n = \(\frac{n}{2}\) [2a + (n – 1) d]
S¹⁶ = \(\frac{16}{2}\) [2 (1) + (16 – 1) \(\frac{1}{2}\)].

Case II:
Putting d = – \(\frac{1}{2}\) in (1), we get,
When d = – \(\frac{1}{2}\)
a + 4 (-\(\frac{1}{2}\)) = 3
a – 2 = 3
or a = 3 + 2 = 5
Using formula,
S_n = \(\frac{n}{2}\) [2a + (n – 1)d]
S₁₆ = \(\frac{16}{2}\) [2(5) + (16 – 1) (-\(\frac{1}{2}\))]
= 8[10 – \(\frac{15}{2}\)]
= 8 \(\left[\frac{20-15}{2}=\frac{5}{2}\right]\)
S₁₆ = 20.

Question 3.
A ladder has rungs 25 cm apart (see fig.) The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and bottom rungs are 2 latex]\frac{1}{2}[/latex] m apart, what is the length of the wood required for the rungs?

[Hint: Number of rungs = \(\frac{250}{25}\) + 1]
Solution:
Total length of rungs = 2 \(\frac{1}{2}\) m = \(\frac{5}{2}\) m
= (\(\frac{5}{2}\) × 100) cm = 250 cm
Length of each rung = 25 cm
∴ Number of rungs = \(\frac{\text { Total length of rungs }}{\text { Length of each rung }}\) + 1
= \(\frac{250}{25}\) + 1 = 10 + 1 = 11
Length of first rung =45 cm
Here a = 45; l = 25; n = 11
Length of the wood for rungs
= S₁₁
= \(\frac{n}{2}\) [a + l]
= \(\frac{11}{2}\) [45 + 25]
= \(\frac{1}{2}\) × 70
= 11 × 35 = 385
Hence, length of the wood for rungs has 385 cm.

Question 4.
The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it and find this value of x.
[Hint: S_{x – 1} = S₄₉ – S₁]
Solution:
Let ‘x’ denotes the number of any house.
Here a = T₁ = 1 ;d = 1
According to question,
S_{x – 1} = S₄₉ – S_x
= \(\frac{x-1}{2}\) [2 (1) + (x – 1 – 1) (1)]
= \(\frac{49}{2}\) [1 + 49] – \(\frac{x}{2}\) [2 (1) + (x – 1) (1)]
[Using S_n = \(\frac{n}{2}\) [2a + (n – 1) d] and S_n = \(\frac{n}{2}\) (a + l) ]
or \(\frac{x-1}{2}\) [2 + x – 2] = \(\frac{49}{2}\) (50) – \(\frac{x}{2}\) [2 + x – 1]
or \(\frac{x(x-1)}{2}=49(25)-\frac{x(x+1)}{2}\)
or \(\frac{x}{2}\) [x – 1 + x + 1] = 1225
\(\frac{x}{2}\) × 2x = 1225
or x² = 1225
or x = 35.

Question 5.
A small terrace at a football ground comprises of 15 step each of which is 50 m long and built of solid concrete. Each step has a rise of \(\frac{1}{4}\) m and a tread of \(\frac{1}{2}\) m (see fig.) Calculate the total volume of concrete required to build the terrace.
[Hint: Volume of concrete required to build of the first step \(\frac{1}{4}\) × \(\frac{1}{2}\) × 50 m³].

Solution:
Volume of concrete required to build the first step \(\frac{1}{4}\) × \(\frac{1}{2}\) × 50 m³]
= [latex]\frac{25}{4}[/latex] m³
Volume of concrete required to build the second step = [\(\frac{25}{4}\) × \(\frac{1}{2}\) × 50] m³

= \(\frac{75}{2}\) m³
Volume of concrete required to build the third step = [\(\frac{3}{4}\) × \(\frac{1}{2}\) × 50] m³ and so on upto 15 steps.

Here a = T₁ = \(\frac{25}{4}\);
T₂ = \(\frac{25}{2}\);
T₃ = \(\frac{75}{4}\); and n = 15.
d = T₂ – T₁ = \(\frac{25}{2}\) – \(\frac{25}{4}\)
= \(\frac{50-25}{4}\) = \(\frac{25}{4}\).

Total volume of concrete required to buld the terrace = S₁₅
= \(\frac{n}{2}\) [2a + (n – 1)d]
= \(\frac{15}{2}\left[2\left(\frac{25}{4}\right)+(15-1) \frac{25}{4}\right]\)
= \(\left[\frac{25}{4} \times \frac{14 \times 25}{4}\right]\)
= \(\frac{15}{2}\left[\frac{25}{2} \times \frac{175}{2}\right]\)
= \(\frac{15}{2} \times \frac{200}{2}\) = 750
Hence, total volume of concrete required to build the terrace is 750 m³.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 5 Arithmetic Progressions Ex 5.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

Question 1.
Find the sum of the following APs:
(i) 2, 7, 12, … to 10 terms.
(ii) – 37, – 33, – 29, ………….. to 12 terms.
(iii) 0.6, 1.7, 2.8, … to 100 terms.
(iv) \(\frac{1}{5}\), \(\frac{1}{12}\), \(\frac{1}{10}\), ………… to 11 terms.
Solution:
(i) Given AP. is 2, 7, 12, …
Here a = 2, d = 7 – 2 = 5 and n = 10
S_n = \(\frac{n}{2}\) [2a + (n – 1) d]
∴ S₁₀ = \(\frac{10}{2}\) [2 × 2 + (10 – 1)5]
= 5 [4 + 45] = 245.

(ii) Given A.P. is – 37, – 33, – 29…
Here a = -37, d = – 33 + 37 = 4 and n = 12
S_n = \(\frac{n}{2}\) [2a + (n – 1) d]
∴ S₁₂ = [2(-37) + (12 – 1)4]
= 6 [- 74 + 44] = – 180.

(iii) Given A.P. is 0.6, 1.7. 2.8.
Here a = 0.6, d = 1.7 – 0.6 = 1.1 and n = 100
S_n = \(\frac{n}{2}\) [2a+(n – 1) d]
∴ S₁₀₀ = \(\frac{100}{2}\) [2(0.6) + (100 – 1) 1.1]
= 50 [1.2 + 108.9] = 5505.

(iv) Given AP. is \(\frac{1}{5}\), \(\frac{1}{12}\), \(\frac{1}{10}\), ………… to 11 terms.
Here a = \(\frac{1}{5}\), d = \(\frac{1}{5}\) and n = 11
S_n = \(\frac{n}{2}\) [2a +(n – 1)d]
∴ S₁₁ = \(\frac{11}{2}\left[2\left(\frac{1}{15}\right)+(11-1) \frac{1}{60}\right]\)

= \(\frac{11}{2}\left[\frac{2}{15}+\frac{10}{60}\right]=\frac{11}{2}\left[\frac{2}{15}+\frac{1}{6}\right]\)

= \(\frac{11}{2}\left[\frac{4+5}{30}=\frac{9}{30}\right]=\frac{33}{20}\).

Question 2.
Find the sums given below:
(i) 7+ 10\(\frac{1}{2}\) + 14 + …………… + 84
(ii) 34 + 32 + 30 + ………. + 10
(iii) – 5 + (- 8) + (- 11) +………+ (- 230)
Solution:
(i) Given A.P. is
7 + 10\(\frac{1}{2}\) + 14 + ………….. + 84
Here a = 7, d = 10\(\frac{1}{2}\) – 7 = \(\frac{21}{2}\) – 7
= \(\frac{21-14}{2}=\frac{7}{2}\)
and l = T_n = 84
or a + (n – 1) d = 84
or 7 + (n – 1) \(\frac{7}{2}\) = 84
or (n – 1) \(\frac{7}{2}\) = 84 – 7 = 77
or n – 1 = 77 × \(\frac{2}{7}\) =22
n = 22 + 1 = 23
∵ S_n = \(\frac{n}{2}\) [a + l]
Now, S₂₃ = \(\frac{23}{2}\) [7 + 84]
= \(\frac{23}{2}\) × 91 = \(\frac{2093}{2}\).

(ii) Given A.P. is 34 + 32 + 30 + …………… + 10
Here a = 34, d = 32 – 34 = -2
l = T_n = 10
a + (n – 1) d = 10
or 34 + (n – 1) (- 2) = 10
or – 2(n – 1) = 10 – 34 = -24
or n – 1 = 12
n = 12 + 1 = 13
[∵ S_n = \(\frac{n}{2}\) [a + l]]
Now, S₁₃ = \(\frac{13}{2}\) [34 + 10]
= \(\frac{23}{2}\) × 44
= 13 × 22 = 286.

(iii) Give A.P. is – 5 + (- 8) + (- 11) + ……………. + (- 230)
Here a = – 5, d = – 8 + 5 = – 3
and l = T_n = – 230
a + (n – 1) d = – 230
or – 5 + (n – 1) (- 3) = – 230
or – 3(n – 1) = – 230 + 5 = – 225
or n – 1 = \(\frac{225}{3}\) = 75
or n = 75 + 1 = 76
Now, S₇₆ = \(\frac{76}{2}\) [- 5 + (- 23o)]
= 38 (- 235) = – 8930.

Question 3.
In an AP:
(i) given a = 5, d = 3, a_n = 50 find n and S_n.
(ii) given a = 7. a₁₃ = 35, find d and S₁₃.
(iii) given a₁₂ = 37, d = 3. find a and S₁₂.
(iv) given a₃ = 15, S₁₀ = 125, find d and a₁₀.
(y) given d = 5, S₉ = 75, find a and a₉.
(vi) given a = 2, d = 8, S_n = 90, find n and a_n.
(vii) given a = 8, a_n = 62, S_n = 210, find n and d.
(viii)given a_n = 4, d = 2, S_n = -14, find n and a.
(ix) given a = 3, n = 8, S = 192, find d.
(x) given l = 28, S = 144, and there are total 9 terms. Find a.
Solution:
(i) Given a = 5, d = 3, a_n = 50
a_n = 50
a + (n – 1) d = 50
or 5 + (n – 1) 3 = 50
or 3 (n – 1) = 50 – 5 = 45
or n – 1 = \(\frac{45}{3}\) = 15
or n = 15 + 1 = 16
Now, S_n = latex]\frac{n}{2}[/latex] [a + l]
= latex]\frac{16}{2}[/latex] [5 + 50] = 8 × 55 = 440.

(ii) Given a = 7, a₁₃ = 35
a₁₃ = 35
a + (n – 1) d = 35
or 7 + (13 – 1) d = 35
or 12 d = 35 – 7 = 28
or d = \(\frac{7}{3}\).
∵ [S_n = latex]\frac{n}{2}[/latex] [a + l]
Now, S₁₃ = \(\frac{13}{2}\) [7 + 35]
= \(\frac{13}{2}\) × 42 = 273

(iii) Given a₁₂ = 37, d = 3
∵ a₁₂ = 37
a + (n – 1) d = 37
or a + (12 – 1) 3 = 37
a + 33 = 37
a = 37 – 33 = 4
∵ [S_n = latex]\frac{n}{2}[/latex] [a + l]
Now, S₁₂ = \(\frac{13}{2}\) [4 + 37]
= 6 × 41 = 246.

(iv)Given a₃ = 15, S₁₀ = 125
∵ a₃ = 15
a + (n – 1) d = 15
a + (3 – 1) d = 15
or a + 2d = 15 …………….(1)
∵ S₁₀ = 125
∵ [S_n = latex]\frac{n}{2}[/latex] [2a + (n – 1)d]
\(\frac{10}{2}\) [2a + (10 – 1) d] = 125
or 5[2a + 9d] = 125
Note this
or 2a + 9d = \(\frac{125}{5}\) = 25
or 2a + 9d = 25 …………(2)
From (1), a = 15 – 24 …………..(3)
Substitute this value of (a) in (2i, we c1
2(15 – 2d) + 9d = 25
or 30 – 4d + 9d = 25
5d = 25 – 30
or d = \(\frac{-5}{5}\) = -1
Substitute this value of d in (3), we get
a = 15 – 2(- 1)
a = 15 + 2 = 17
Now, a₁₀ = 17 + (10 – 1)(- 1)
∵ T_n = a + (n – 1) d = 17 – 9 = 8.

(v) Given d = 5, S₉ = 75
∵ S₉ = 75
∵ [S_n = \(\frac{n}{2}\) [2a + (n – 1)d]
\(\frac{9}{2}\) [2a + 40] = 75
or [2a + 40] = \(\frac{50}{3}\)
or 2a = \(\frac{50}{3}\) – 40
or a = \(-\frac{70}{3} \times \frac{1}{2}\)
or a = – \(\frac{35}{3}\)
Now, a₉ = a + (n – 1) d
= – \(\frac{35}{3}\) + (9 – 1)⁰ × 5
= – \(\frac{35}{3}\) + 4o = \(\frac{-35+120}{3}\)
a₉ = \(\frac{85}{3}\).

(vi) Given a = 2, d = 8, S_n = 90
∵ S_n = 90
\(\frac{n}{2}\) [2a + (n – 1)d] = 9o
or \(\frac{n}{2}\) [2 × 2 + (n – 1) 8] = 90
or n [2 + 4n – 4] = 90
or n (4n – 2) = 90
or 4n² – 2n – 90 = 0
or 2n² – 10n + 9n – 45 = 0
S = – 2
P = – 45 × 2 = – 90
or 2n [n – 5] + 9(n – 5) = 0
or (2n + 9) (n – 5) = 0
Either 2n + 9 = 0 or n – 5 = 0
Either n = – \(\frac{9}{2}\) or n = 5
∵ n cannot be negative so reject n = – \(\frac{9}{2}\)
∴ n = 5
Now, a_n = a₅ = a + (n – 1) d
= 2 + (5 – 1) 8 = 2 + 32 = 34.

(vii) Given a = 8, a_n = 62, S_n = 210
∵ S_n = 210
\(\frac{n}{2}\) [a + a_n] = 210
or \(\frac{n}{2}\) [8 + 62] = 210
or \(\frac{n}{2}\) × 70 = 210
or n = \(\frac{210}{35}\) = 6
Now, a_n = 62
[∵ T_n = a + (n – 1) d]
or 5d = 62 – 8 = 54
or d = \(\frac{54}{5}\).

(viii) Given a_n = 4, d = 2, S_n = – 14
∵ a_n = 4
a + (n – 1) d = 4
or a + (n – 1)2 = 4
or a + 2n – 2 = 4
or a = 6 – 2n ……………(1)
and S_n = – 14
\(\frac{n}{2}\) [a + a_n] = – 14
or \(\frac{n}{2}\) [6 – 2n +4] = – 14 (Using (1))
or \(\frac{n}{2}\) [10 – 2n] = – 14
or 5n – n² + 14 = 0
or n² – 5n – 14 = 0
S = – 5
P = 1 × – 14 = – 14
or n² – 7n + 2n – 14 = 0
or n(n – 7) + 2 (n – 7) = 0
or (n – 7) (n + 2) = 0
either n – 7 = 0
or n + 2 = 0
n = 7 or n = -2
∵ n cannot be negative so reject n = – 2
∴ n = 7
Substitute this value of n in (1), we get
a = 6 – 2 × 7
a = 6 – 14 = – 8.

(ix) Given a = 3, n = 8, S = 192
∵ S = 192
S₈ = 192 [∵ n = 8]
∵ S_n = \(\frac{n}{2}\) [2a + (n – 1)d]
or \(\frac{8}{2}\) [2 × 3 + (8 – 1) d] = 192
or 4 [6 + 7d] = 192
6 + 7d = \(\frac{192}{4}\) = 48
or 7d = 48 – 6 = 42
or d = \(\frac{42}{6}\) = 6.

(x) Given l = 28, S = 144 and there are total 9 terms
∴ n = 9; l = a₉ = 28; S₉ = 144
∵ a₉ = 28
or a + (9 – 1) d = 28
∵ [a_n = T_n = a + (n – 1) d ]
or a + 8d = 28 …………….(1)
and S₉ = 144
∵ [S_n = \(\frac{n}{2}\) [a + a_n]]
\(\frac{9}{2}\) [a + 28] = 144
or a + 28 = \(\frac{144 \times 2}{9}\) = 32
a = 32 – 28 = 4.

Question 4.
How many terms of the A.P : 9, 17, 5… must be taken to give a sum of 636?
Solution:
Given A.P. is 9, 17, 25, …………
Here a = 9, d = 17 – 9 = 8
But S_n = 636
\(\frac{n}{2}\) [2a + (n – 1) d] = 636
or \(\frac{n}{2}\) [2(9) + (n – 1) 8] = 636
or \(\frac{n}{2}\) [18 + 8n – 8] = 636
or n [4n + 5] = 636
or 4n² + 5n – 636 = 0
a = 4, b = 5, c = – 636
D = (5) – 4 × 4 × (- 636)
= 25 + 10176 = 10201
∴ n = \(\frac{-b \pm \sqrt{\mathrm{D}}}{2 a}\)

= \(\frac{-5 \pm \sqrt{10201}}{2 \times 4}=\frac{-5 \pm 101}{8}\)

= \(\frac{-106}{8} \text { or } \frac{96}{8}\)

= \(-\frac{53}{4}\) or 12
∴ n cannot be negative so reject n = \(-\frac{53}{4}\)
∴ n = 12
Hence, sum of 12 terms of given A.P. has sum 636.

Question 5.
The first term ofan AP is 5, the last term is 45 and the sum Is 400. Find the number of terms and the common difference.
Solution:
Given that a = T₁ = 5; l = a_n = 45
and S_n = 400
∴ T_n = 45
a + (n – 1) d = 45
or 5 + (n – 1) d = 45
or (n – 1) d = 45 – 5 = 40
or (n – 1) d = 40 ……………….(1)
and S_n = 400
\(\frac{n}{2}\) [a + a_n] = 400
or \(\frac{n}{2}\) [5 + 45] = 400
or 25 n = 400
or n = \(\frac{400}{25}\) = 16
Substitute this value of n in (1), we get
(16 – 1) d = 40
or 15d = 40
d = \(\frac{40}{15}\) = \(\frac{8}{3}\)
Hence, n = 16 and d = \(\frac{8}{3}\)

Question 6.
The first and last terms of an AP are 17 and 350 respectively. 1f the common difference is 9, how many terms are there
and what is their sum?
Solution:
Given that a = T₁ = 17;
l = a_n = 350 and d = 9
∵ l = a_n = 350
a + (n – 1) d = 350
17 + (n – 1) 9 = 350
or 9 (n – 1) = 350 – 17 = 333
or n – 1 = \(\frac{333}{9}\) = 37
n = 37 + 1 = 38
Now, S₃₈ = \(\frac{n}{2}\) [a + l]
= \(\frac{38}{2}\) (17 + 350]
= 19 × 367 = 6973.
Hence, sum of 38 terms of given A.P. arc 6973.

Question 7.
Find the sum of first 22 terms of an AP. in which d = 7 and 22nd term is 149.
Solution:
Given that d = 7; T₂₂ = 149 and n = 22
∵ T₂₂ = 149
a + (n – 1) d = 149
or a + (22 – 1) 7 = 149
or a + 147 = 149
or a = 149 – 147 = 2
Now, S₂₂ = [a + T₂₂]
= \(\frac{22}{2}\) [2+ 149] = 11 × 151 = 1661
Hence, sum of first 22 temis of given A.P. is 1661.

Question 8.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:
Let ‘a’ and ‘d’ be fïrst term and common difference
Given that T₂ = 14; T₃ = 18 and n = 51
∵ T₂ = 14
a + (n – 1) d = 14
a + (2 – 1)d = 14
or a + d = 14
a = 14 – d …………….(1)
and T₃ = 18 (Given)
a + (n – 1) d = 18
a + (3 – 1) d = 18
or a + 2d = 18
or 14 – d + 2d = 18
or d = 18 – 14 = 4
or d = 4
Substitute this value of d in (1), we get
a = 14 – 4 = 10
Now, S₅₁ = \(\frac{n}{2}\) [2a + (n – 1) d]
= \(\frac{51}{2}\) [2 × 10 + (51 – 1) 4]
= \(\frac{51}{2}\) [2o + 2oo]
= \(\frac{51}{2}\) × 220 = 51 × 110 = 5610
Hence, sum of first 51 terms of given A.P. is 5610.

Question 9.
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of given A.P.
According to 1st condition
S₇ = 49
\(\frac{n}{2}\) [2a + (n – 1) d] = 49
or \(\frac{7}{2}\) [2a + (7 – 1) d] = 49
or \(\frac{7}{2}\) [2a + 6d] = 49
or a + 3d = 7
or a = 7 – 3d
According to 2nd condition
S₁₇ = 289
\(\frac{n}{2}\) [2a+(n – 1)d]=289
\(\frac{17}{2}\) [2a + (17 – 1) d] = 289
a + 8d = \(\frac{289}{17}\) = 17
Substitute the value of a from (1), we get
7 – 3d + 8d = 17
5d = 17 – 7 = 10
d = \(\frac{10}{5}\) = 2
Substitute this value of d in (1), we get
a = 7 – 3 × 2
a = 7 – 6 = 1
Now, S_n = \(\frac{n}{2}\) [2a + (n – 1) d]
= \(\frac{n}{2}\) [2 × 1 + (n – 1) × 2] = n [I + n – 1] = n × n = n²
Hence, sum of first n terms of given A.P. is n².

Question 10.
Show that a₁, a₂, ……., a_n, … form an AP where is defined as below.
(i) a_n = 3 + 4n
(ii) a_n = 9 – 5n
Also find the sum of the first 15 terms In each case.
Solution:
(i) Given that a_n = 3 + 4n ………..( 1)
Putting the different values of n in(1), we get
a₁ = 3 + 4(1) = 7;
a₂ = 3 + 4 (2) = 11
a₃ = 3 + 4 (3) = 15, …………
Now, a₂ – a₁ = 11 – 7
and a₃ – a₂ = 15 – 11 = 4
a₂ – a₁ = 11 – 7 = 4
and a₃ – a₂ = 4 = d(say)
∵ given sequence form an A.P.
Here a = 7, d = 4 and n = 15
∴ S₁₅ = \(\frac{n}{2}\) [2a + (n – 1)d]
= \(\frac{15}{2}\) [2(7) + (15 – 1) 4]
= \(\frac{15}{2}\) [14 + 56] = \(\frac{15}{2}\) × 70
= 15 × 35 = 525.

(ii) Given that a_n = 9 – 5n ……………….(1)
Putting the different values of n is (1), we get
a₁ = 9 – 5(1) = 4;
a₂ = 9 – 5(2) = -1;
a₃ = 9 – 5(3) = -6.
Now, a₂ – a₁ = – 1 – 4 = – 5
and a₃ – a₂ = – 6 + 1 = – 5
∵ a – a₁ = a₃ – a₂ = – 5 = d (say)
∴ given sequence form an A.P.
Here a = 4, d = – 5 and n = 15
∴ S₁₅ = \(\frac{n}{2}\) [2a + (n – 1) d]
= \(\frac{15}{2}\) [2(4) + (15 – 1) (-5)]
= \(\frac{15}{2}\) [8 – 70]
= \(\frac{15}{2}\) (-62) = – 465

Question 11.
If the sum of the first n terms of an AP is 4n – n², what is the first term (that is S₁) ? What is the sum of two terms? What is the second term ? Similarly, find the 3rd, the 10th and the nth terms.
Solution:
Given that, sum of n terms of an A.P. are
S_n = 4n – n²
Putting n = 1 in (1), we get
S₁ = 4(1) – (1)² = 4 – 1
S₁ = 3
∴ a = T₁ = S₁ = 3
Putting n = 2, in (1), we get
S₂ = 4(2) – (2)² = 8 – 4
S₂ = 4
or T₁ + T₂ = 4
or 3 + T₂ = 4
or T₂ = 4 – 3 = 1
Putting n = 3 in (1), we get
S₃ =4(3) – (3)² = 12 – 9
S₃ = 3
or S₂ + T₃ = 3
or 4+ T₃ = 3
or T₃ = 3 – 4 = – 1
Now, d = T₂ – T₁
d = 1 – 3 = -2
∴ T₁₀ = a + (n – 1) d
= 3 + (10 – 1) (- 2)
T₁₀ = 3 – 18 = – 15
and T_n = a + (n – 1) d
= 3 + (n – 1) (- 2)
= 3 – 2n + 2
T_n = 5 – 2n.

Question 12.
Find the sum of the first 40 positive integers divisible by 6.
Solution:
Positive integers divisible by 6 are 6, 12, 18, 24, 30, 36 42, …
Here a = T₁ = 6, T₂ = 12, T₃ = 18, T₄ = 24
T₂ – T₁ = 12 – 6 = 6
T₃ – T₂ = 18 – 12 = 6
T₄ – T₃ = 24 – 18 = 6
∵ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = 6 = d (say)
Using formula, S_n = \(\frac{n}{2}\) [2a + (n – 1) d]
S₄₀ = \(\frac{40}{2}\) [2(6) + (40 – 1)6].
= 20 [12 + 234]
= 20 (246) = 4920
Hence, sum of first 40 positive integers divisible by 6 is 4920.

Question 13.
Find the sum of first 15 multiples of 8.
Solution:
Multiples of 8 are 8, 16, 24, 32, 40, 48, …………..
Here a = T₁ = 8; T₂ = 16; T₃ = 24 ; T₄ = 32
T₂ – T₁ = 16 – 8= 8
T₃ – T₂ = 24 – 16=8
T₂ – T₁ = T₃ – T₂ = 8 = d(say)
Ùsing formula. S_n = [2a + (n – 1) d}
S₁₅ = \(\frac{15}{2}\) [2(8) + (15 – 1) 8]
= \(\frac{15}{2}\) [16 + 112]
= \(\frac{15}{2}\) × 128 = 960
Hence, sum of first 15 multiples of 8 is 960.

Question 14.
Find the sum of the odd numbers between 0 and 50.
Solution:
Odd numbers between 0 and 50 are 1, 3, 5, 7, 9, …………, 49
Here a = T₁ = 1; T₂ = 3; T₃ = 5 ; T₄ = 7
and l = T_n = 49
T₂ – T₁ = 3 – 1 = 2
T₃ – T₂ = 5 – 3 = 2
∵ T₂ – T₁ = T₃ – T₂ = 2 = d (say)
Also, l = T_n = 49
a + (n – 1) d = 49
1 + (n – 1) 2 = 49
or 2 (n – 1) = 49 – 1 = 48
n – 1 = \(\frac{48}{2}\) = 24
n = 24 + 1 = 25.
Using formula, S_n = \(\frac{n}{2}\) [2a + (n – 1) d]
S₂₅ = \(\frac{25}{2}\) [2(1) + (25 – 1)2]
= \(\frac{25}{2}\) [2 + 48]
= \(\frac{25}{2}\) × 50 = 625
Hence, sum of the odd numbers between 0 and 50 are 625.

Question 15.
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹ 200 for
the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc., the penalty for each succeeding day being ₹ 50 more than for the preceding day, How much money the contractor has to pay as penalty, if he hasdelayed the work by 30 days?
Solution:
Penalty (cost) for delay of one, two, third day are ₹ 200, ₹ 250, ₹ 300
Now, penalty increase with next day with a difference of ₹ 50.
∴ Required A.P. are ₹ 200, ₹ 250, ₹ 300, ₹ 350, …
Here a = T₁ = 200; d = ₹ 50 and n = 30
Amount of penalty gives after 30 days
= S₃₀
= \(\frac{n}{2}\) [2a + (n – 1) d]
= \(\frac{30}{2}\) [2(2oo) + (30 – 1) 50)
= 15 [400 + 1450]
= 15 (1850) = 27750
Hence, ₹ 27350 pay as penalty by the contractor if he has delayed the work 30 days.

Question 16.
A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If
each prize is ₹ 20 less than its preceding prize, find the value of each of the prizes.
Solution:
Let amount of prize given to 1st student = ₹ x
Amount of prize given to 2nd student = ₹ (x – 20)
Amount of prize given to 3rd student = ₹ [x – 20 – 20] = ₹ (x – 40)
and so on.
∴ Required sequence are ₹ x, ₹ (x – 20), ₹ (x – 40), … which form on A.P. with
a = ₹ x, d = – 20 and n = 7
Using formula, S_n = \(\frac{n}{2}\) [2a + (n – 1) d]
S₇ = \(\frac{7}{2}\) [2(x) + (7 – 1) (- 20)]
S₇ = \(\frac{7}{2}\) [2x – 120] = 7 (x – 60)
According to question,
7 (x – 60) = 700
x – 60 = 7 = 100
x – 60 = 7 = 100
x = 100 + 60
x = 160
Hence, 7 prizes are ₹ 160, ₹ 140, ₹ 120, ₹ 100, ₹ 80, ₹ 60, ₹ 40.

Question 17.
In a school, student thought of planting trees in and around the school to reduce air pollution. It was decided that
number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g. – a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
Solution:
Number of trees planted by three sections of class I = 3 × 1 = 3
Number of trees planted by three sections of class II = 3 × 2 = 6
Number of trees planted by three sections of class III = 3 × 3 = 9
…………………………………………………………………..
…………………………………………………………………..
……………………………………………………………………
Number of trees planted by three sections of class XII = 3 × 12 = 36
:. Required A.P. are 3, 6, 9, …………., 36
Here a = T₁ = 3; T₂ = 6; T₃ = 9
and l = T_n = 36; n = 12
d = T₂ – T₁ = 6 – 3 = 3
Total number of trees planted by students
= S₁₂
= \(\frac{n}{2}\) [a + l]
= \(\frac{12}{2}\) [3+ 36]
= 6 × 39 = 234
Hence, 234 trees will be planted by students to reduce air pollution.

Question 18.
A spiral is made up of successive semicircics, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, …. as shown in Fig. What is the total length of such a spiral made up of thirteen consecutive semicircies? (Take π = \(\frac{22}{7}\))
[Hint: Length of successive semicircies is ‘l₁’ ‘l₂’ ‘l₃‘ ‘l₄‘ … wIth centres at A, B, A, B, …, respectively.]

Solution:
Let l₁ = length of first semi circle = πr₁ = π(0.5) = \(\frac{\pi}{2}\)
l₂ = length of second semi circlem = πr₂= π(1) = π
l₃ = length of third semi circle = πr₃ = π(1.5) = \(\frac{3 \pi}{2}\)
and l₄ = length of fourth senil circle = πr₄ = π(2) = 2π and so on
∵ length of each successive semicircle form an A.P.
Here
a = T₁ = \(\frac{\pi}{2}\); T₂ = π;
T₃ = \(\frac{3 \pi}{2}\); T₄ = 2π……… and n = 13
d = T₂ – T₁ = π – \(\frac{\pi}{2}\)
= \(\frac{2 \pi-\pi}{2}=\frac{\pi}{2}\)
Length of whole spiral = S₁₃

Hence, total length of a spiral made up of thirteen consecutive semi circies is 143 cm

Question 19.
200 logs are stacked in the following manner : 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see Fig). In how many rows are the 200 logs placed and how many logs are in the top row?

Solution:
Number of logs in the bottom (1st row) = 20
Number of logs in the 2nd row = 19
Number of log in the 3rd row = 18 and so on.
∴ Number of logs in the each steps form an
Here a = T₁ = 20; T₂ = 19; T₃ = 18…
d = T₂ – T₁ = 19 – 20 = – 1
Let S, denotes the number logs.
Using formula, S_n = \(\frac{n}{2}\) [2a + (n – 1) d]
= \(\frac{n}{2}\) [2 (20) + (n – 1) (-1)]
∴ S_n = \(\frac{n}{2}\) [40 – n + 1]
= \(\frac{n}{2}\) [41 – n]
According to question,
\(\frac{n}{2}\) [41 – n] = 200
or 41 – n² = 400
or – n² + 41n – 400 = 0
or n² – 41n + 400 = 0
S = – 41, P = 400
or n² – 16n – 25n + 400 = 0
or n (n – 16) – 25 (n – 16)=0
or (n – 16) (n – 25) = 0
Either n – 16 = 0 or n – 25 = 0
Either n = 16 or n = 25
∴ n = 16, 25.
Case I:
When n = 25
T₂₅ = a + (n – 1) d
= 20 + (25 – 1)(- 1)
= 20 – 24 = – 4;
which is impossible
∴ n = 25 rejected.

Case II. When n = 16
T₁₆ = a + (n – 1) d
= 20 + (16 – 1)(- 1)
= 20 – 15 = 5
Hence, there are 16 row and 5 logs are in the top row.

Question 20.
In a potato race a bucket ¡s placed at the starting point which Ls 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see Fig.)

Each competitor starts from the bucket, picks up the nearest potato, runs back with It, drops It in the bucket, runs back to pick up the next potato, runs to the bucket to drop it In, and the continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
(Hint : To pick up the first potato and second potato, the distance run is [2 × 5 + 2 × (5 + 3)] m)
Solution:
Distance covered to pick up the Ist potato = 2(5) m = 10 m
Distance between successive potato = 3 m
distance covered to pick up the 2nd potato = 2(5 + 3) m = 16 m
Distance covered to pick up the 3rd potato = 2 (5 + 3 + 3) m = 22 m
and this process go on. h is clear that this situation becomes an A.P. as 10 m, 16 m, 22 m, 28 m, …
Here a = T₁ = 10; T₂ = 16; T₃ = 22, …
d = T₂ – T₁ = 16 – 10 = 6 and n = 10
∴ Total distance the competitor has to run = S₁₀
= \(\frac{n}{2}\) [2a + (n – 1) d]
= \(\frac{10}{2}\) [2(10) + (10 – 1) 6]
= 5 [20 + 54]
= 5 × 74 = 370
Hence, 370 m is the total distance run by a competitor.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.2

Q.uestion 1.
Find the co-ordinates of the point which divides the join (- 1, 7) and (4, – 3) in the ratio 2 : 3.
Solution:
Let required point be P (x, y) which divides the join of given points A (- 1, 7)
and B (4, – 3) in the ratio of 2 : 3.
(-1, 7) (x, y) (4, – 3)

∴ x = \(\frac{2 \times 4+3 \times-1}{2+3}=\frac{8-3}{5}=\frac{5}{5}=1\)

and y = \(\frac{2 \times-3+3 \times 7}{2+3}=\frac{-6+21}{5}=\frac{15}{5}=3\)
Hence, required point be (1, 3).

Question 2.
Find the co-ordinates of the points of trisection of the line segment joining (4, – 1) and (2, – 3).
Solution:
Let P (x₁, y₁) and Q (x₂, y₂) be the required points which trisect the line segment joining A (4, – 1)and B (- 2, – 3) i.e., P(x₁, y₁) divides AB in ratio 1: 2 and Q divides AB in ratio 2 : 1.

∴ x₁ = \(\frac{1 \times-2+2 \times 4}{1+2}=\frac{-2+8}{3}=\frac{6}{3}=2\)

and y₁ = \(\frac{1 \times-3+2 \times-1}{1+2}=\frac{-3-2}{3}=-\frac{5}{3}\)
∴ P(x₁, y₁) be (2, \(-\frac{5}{3}\))

Now, x₂ = \(\frac{2 \times-2+1 \times 4}{2+1}\)
= \(\frac{-4+4}{3}\) = 0

y₂ = \(\frac{2 \times-3+1 \times-1}{2+1}=\frac{-6-1}{3}=-\frac{7}{3}\)

∴ Q(x₂, y₂) be (0, \(-\frac{7}{3}\))
Hence, required points be (2, \(-\frac{5}{3}\)) and (0, \(-\frac{7}{3}\)).

Question 3.
To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown in fig. Niharika runs \(\frac{1}{4}\) th the distance AD on the 2nd line and posts a green flag.

Preet runs \(\frac{1}{5}\) th the distance AD on the eighth line and posts a red flag. What is the distance betweenboth the flags? If Rashmi has to post a blue flag exactly half way between the line (segment) joining the two flags, where should she post her flag?

Solution:
In the given figure, we take A as origin. Taking x-axis along AB and y-axis along AD.
Position of green flag = distance covered by Niharika
= Niharika runs \(\frac{1}{4}\)th distance AD on the 2nd line
= \(\frac{1}{4}\) × 100 = 25 m
∴ Co-ordinates of the green flag are (2, 25)
Now, position of red flag = distance covered by Preet = Preet runs \(\frac{1}{5}\)th the distance AD on the 8th line
= \(\frac{1}{5}\) × 100 = 20 m.
Co-ordinates of red flag are (8, 20)
∴ distance between Green and Red flags = \(\sqrt{(8-2)^{2}+(20-25)^{2}}\)
= \(\sqrt{36+25}=\sqrt{61}\) m.
Position of blue flag = mid point of green flag and red flag
= \(\left(\frac{2+8}{2}, \frac{25+20}{2}\right)\)
= (5, 22.5).
Hence, blue flag is in the 5th line and at a distance of 22.5 m along AD.

Question 4.
Find the ratio in which (he segment joining the points (- 3, 10) and (6, – 8) is divided by (- 1, 6).
Solution:
Let point P (- 1, 6) divides the line segment joining the points A (- 3, 10) and B (6, – 8) the ratio K : 1.

∴ -1 = \(\frac{6 \times \mathrm{K}-3 \times 1}{\mathrm{~K}+1}\)
or – K – 1 = 6K – 3
or – K – 6K = – 3 + 1
or – 7K = – 2
K : 1 = \(\frac{2}{7}\) : 1 = 2 : 7
Hence, required ratio is 2 : 7.

Question 5.
Find the ratio in which the line segment joining A (1, – 5) and B (- 4, 5) is divided by the x-axis. Also find the co
ordinates of the point of division.
Solution:
Let required point on x-axis is P (x, 0) which divides the line segment joining the points A (1, – 5) and B (- 4, 5) in the
ratio K : 1.

Consider, y co-ordinates of P ¡s:
0 = \(\frac{5 \times \mathrm{K}+(-5) \times 1}{\mathrm{~K}+1}\)

or 0 = \(\frac{5 \mathrm{~K}-5}{\mathrm{~K}+1}\)
or 5K – 5 = 0
or 5K = 5
or K = \(\frac{5}{5}\) = 1
∴ Required ratio is K : 1 = 1 : 1.
Now, x co-ordinate of P is:
x = \(\frac{-4 \times K+1 \times 1}{K+1}\)
Putting the value of K = 1, we get:
x = \(\frac{-4 \times 1+1 \times 1}{1+1}=\frac{-4+1}{2}\)
x = \(-\frac{3}{2}\)
Hence, required point be (\(-\frac{3}{2}\), 0).

Question 6.
If (1, 2); (4, y); (x, 6) and (3, 5)are the vertices of a parallelogram taken in order, find x and y.
Solution:
Let points of parallelogram ABCD are A (1, 2) (4, y) ; C (x, 6) and D (3, 5)
But diagonals of a || gm bisect each other.
Case I. When E is the mid point of A (1, 2) and C (x, 6)
∴ Co-ordinates of E are:
E = \(\left(\frac{x+1}{2}, \frac{6+2}{2}\right)\)
E = (\(\frac{x+1}{2}\), 4) …………..(1)

Case II. When E is the mid point B (4, y) and D (3, 5)
∴ Co-ordinates of E are:

E = \(\left(\frac{3+4}{2}, \frac{5+y}{2}\right)\)

E = \(\left(\frac{7}{2}, \frac{5+y}{2}\right)\) …………….(2)
But values of E in (1) and (2) are same, so comparing the coordinates, we get
\(\frac{x+1}{2}=\frac{7}{2}\)
or x + 1 = 7
or x = 6.

and 4 = \(\frac{5+y}{2}\)
or 8 = 5 + y
or y = 3
Hence, values of x and y are 6 and 3.

Question 7.
Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4).
Solution:
Let, coordinates of A be (x, y). But, centre is the’ niij ioint of the vertices of the diameter.

∴ O is the mid point of A(x, y) and B(1, 4)
∴ \(\left(\frac{x+1}{2}, \frac{y+4}{2}\right)\) = (2, -3)
On comparing, we get
\(\frac{x+1}{2}\)
or x + 1 = 4
or x = 3

and \(\frac{y+4}{2}\) = – 3
or y + 4 = – 6
or y = – 10
Hence, required point A be (3, – 10).

Question 8.
If A and B are (- 2, – 2) and (2, – 4) respectively, find the coordinates of P such that AP = AB and P lies ¡n the line segment AB.
Solution:
Let required point P be (x, y)
Also AP = \(\frac{3}{7}\) AB …(Given)
But, PB = AB – AP
= AB – \(\frac{3}{7}\) AB = \(\frac{7-3}{7}\) AB
PB = \(\frac{4}{7}\) AB
∴ \(\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\frac{3}{7} \mathrm{AB}}{\frac{4}{7} \mathrm{AB}}=\frac{3}{4}\).

∴ P divides given points A and B in ratio 3 : 4.
Now,
x = \(\frac{3 \times 2+4 \times-2}{3+4}\)

or x = \(\frac{6-8}{7}=-\frac{2}{7}\)

and y = \(\frac{3 \times-4+4 \times-2}{3+4}\)
= \(\frac{-12-8}{7}=-\frac{20}{7}\)

Hence, coordinates of P be (\(-\frac{2}{7}\), \(-\frac{20}{7}\)).

Question 9.
Find the coordinates of the points which divides the line segment joining A (- 2, 2) and B (2, 8) into four equal parts.
Solution:
Let required points are C, D and E which divide the line segment joming the points A (- 2, 2) and B (2, 8) into four equal parts. Then D is mid point of A and B ; C is the mid point of A and D ; E is the mid point of D and B such that
AC = CD = DE = EB

Now, mid point of A and B (i.e., Coordinates of D)
= \(\left(\frac{-2+2}{2}, \frac{2+8}{2}\right)\) = (0, 5)

Mid point of A and D (i.e., Coordinates of C)
= \(\left(\frac{-2+0}{2}, \frac{2+5}{2}\right)=\left(-1, \frac{7}{2}\right)\)

Mid point of D and B (i.e., Coordinates of E)
= \(\left(\frac{2+0}{2}, \frac{8+5}{2}\right)=\left(1, \frac{13}{2}\right)\)

Hence, requned points be (0, 5), (-1, \(\frac{7}{2}\)), (1, \(\frac{13}{2}\)).

Question 10.
Find the area of a rhombus if the vertices are (3, 0); (4, 5); (- 1, 4) and(- 2, – 1) taken in order.
[Hint: Areas of a rhombus = \(\frac{1}{2}\) (Product of its diagonals)]
Solution:
Let coordinates of rhombus ABCD are A (3, 0); B(4, 5); C(-1, 4) and D(- 2, – 1).
Diagonal, AC = \(\sqrt{(-1-3)^{2}+(4-0)^{2}}\)
= \(\sqrt{16+16}=\sqrt{32}=\sqrt{16 \times 2}\) = 4√2

and diagonal BD
BD = \(\sqrt{(-2-4)^{2}+(-1-5)^{2}}\)
= \(\sqrt{36+36}=\sqrt{72}=\sqrt{36 \times 2}\) = 6√2.

∴ Area of rhombus ABCD = \(\frac{1}{2}\) × AC × BD
ABCD = [\(\frac{1}{2}\) × 4√2 × 6√2] sq. units
(\(\frac{1}{2}\) × 24 × 2) sq. units
= 24 sq. units
Hence, area of rhombus is 24 sq. units.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 7 Coordinate Geometry Ex 7.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.1

Question 1.
Find the distance between the following pairs of points:
(i) (2, 3); (4, 1)
(ii)(-5, 7); (-1, 3)
(iii) (a, b); (-a, -b).
Solution:
(i) Given points are: (2, 3); (4, 1)
Required distance = \(\sqrt{(4-2)^{2}+(1-3)^{2}}\)
\(\sqrt{4+4}=\sqrt{8}=\sqrt{4 \times 2}\)
= 2√2.

(ii) Given points are: (-5, 7); (-1, 3)
Required distance = \(\sqrt{(-1+5)^{2}+(3-7)^{2}}\)
\(\sqrt{16+16}=\sqrt{32}\)
= \(\sqrt{16 \times 2}\)
= 4√2.

(iii) Given points are: (a, b); (-a, -b)
Required distance = \(\sqrt{(-a-a)^{2}+(-b-b)^{2}}\)
= \(\sqrt{(-2 a)^{2}+(-2 b)^{2}}\)
= \(\sqrt{4 a^{2}+4 b^{2}}\)
= √4 \(\sqrt{a^{2}+b^{2}}\)
= \(2 \sqrt{a^{2}+b^{2}}\)

Question 2.
Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B
discussed in section 7.2.
Solution:
Given points are: A (0, 0) and B (36, 15)
Distance, AB = \(\sqrt{(0-36)^{2}+(0-15)^{2}}\)
\(\sqrt{1296+225}=\sqrt{1521}\) = 39.
According to Section 7.2
Draw the distinct points A (0, 0) and B (36, 15) as shown in figure.

Draw BC ⊥ on X-axis.
Now, In rt. ∠d ∆ACB,
AB = \(\sqrt{\mathrm{AC}^{2}+\mathrm{BC}^{2}}\)
= \(\sqrt{(36)^{2}+(15)^{2}}\)
= \(\sqrt{1296+225}=\sqrt{1521}\)
= 39.
Hence, required distance between points is 39.

Question 3.
Determine if the points (1, 5), (2, 3) and (- 2, – 11) are collinear.
Solution:
Given point are : A (1. 5); B (2.3) and C (- 2, – 11).
AB = \(\sqrt{(2-1)^{2}+(3-5)^{2}}\)
= \(\sqrt{1+4}=\sqrt{5}\)

BC = \(\sqrt{(-2-2)^{2}+(-11-3)^{2}}\)
= \(\sqrt{16+196}=\sqrt{212}\)

CA = \(\sqrt{(1+2)^{2}+(5+11)^{2}}\)
= \(\sqrt{9+256}=\sqrt{265}\)
From above distances, it is clear that sum of any two is not equal to third one.
Hence, given points are not collinear

Question 4.
Check whether (5, – 2); (6, 4) and (7, – 2) are the Vertices of an isosceles triangle.
Solution:
Given points be A (5, – 2); B (6, 4) and C (7, – 2).
AB = \(\sqrt{(5-6)^{2}+(-2-4)^{2}}\)
= \(\sqrt{1+36}=\sqrt{37}\)

BC = \(\sqrt{(6-7)^{2}+(4+2)^{2}}\)
= \(\sqrt{1+36}=\sqrt{37}\)

CA = \(\sqrt{(7-5)^{2}+(-2+2)^{2}}\)
= \(\sqrt{4+0}=2\)
From above discussion, it is clear that AB = BC = √37.
Given points are vertices of an isosceles triangle.

Question 5.
In a classroom, 4 friends are seated at the points A, B, C and D as shown in fig. Champa and Charnel walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square”? Chameli disagrees. Using distance formula, find which of them is correct, and why?

Solution:
In the given diagram, the vertices of given points are : A (3, 4); B (6, 7); C (9, 4) and D (6, 1).
Now,
AB = \(\sqrt{(6-3)^{2}+(7-4)^{2}}\)
= \(\sqrt{9+9}=\sqrt{18}\)

BC = \(\sqrt{(9-6)^{2}+(4-7)^{2}}\)
= \(\sqrt{9+9}=\sqrt{18}\)

CD = \(\sqrt{(6-9)^{2}+(1-4)^{2}}\)
= \(\sqrt{9+9}=\sqrt{18}\)

DA=\(\sqrt{(3-6)^{2}+(4-1)^{2}}\)
= \(\sqrt{9+9}=\sqrt{18}\)

AC = \(\sqrt{(9-3)^{2}+(4-4)^{2}}\)
= \(\sqrt{36+0}=6\)

BD = \(\sqrt{(6-6)^{2}+(1-7)^{2}}\)
= \(\sqrt{0+36}\) = 6
From above discussion, it is clear that
AB = BC = CD = DA = √18 and AC = BD = 6.
ABCD formed a square and Champa is correct about her thinking.

Question 6.
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) ( 1,- 2), (1, 0),(- 1, 2), (- 3, 0)
(ii) ( 3, 5), (3, 1), (0, 3), (- 1, – 4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2).
Solution:
(i) Given points be A (- 1, – 2); B(1, 0); C(- 1, 2) and D(- 3, 0).
AB = \(\sqrt{(1+1)^{2}+(0+2)^{2}}\)
= \(\sqrt{4+4}=\sqrt{8}\)

BC = \(\sqrt{(-1-1)^{2}+(2-0)^{2}}\)
= \(\sqrt{4+4}=\sqrt{8}\)

CD = \(\sqrt{(-3+1)^{2}+(0-2)^{2}}\)
= \(\sqrt{4+4}=\sqrt{8}\)

DA = \(\sqrt{(-1+3)^{2}+(-2+0)^{2}}\)
= \(\sqrt{4+4}=\sqrt{8}\)

AC = \(\sqrt{(-1+1)^{2}+(2+2)^{2}}\)
= \(\sqrt{0+16}=4\)

BD = \(\sqrt{(-3-1)^{2}+(0-0)^{2}}\)
= \(\sqrt{16+0}=4\)

From above discussion, it is clear that
AB = BC = CD = DA = √8 and AC = BD = 4.
Hence, given quadrilateral ABCD is a square.

(ii) Given points be A (- 3, 5); B (3, 1); C (0, 3) and D (- 1,- 4)
AB = \(\sqrt{(-3-3)^{2}+(5-1)^{2}}\)
= \(\sqrt{36+16}=\sqrt{52}=\sqrt{4 \times 13}\)
= 2√13

BC = \(\sqrt{(3-0)^{2}+(1-3)^{2}}\)
= \(\sqrt{9+4}=\sqrt{13}\)

CA = \(\sqrt{(0+3)^{2}+(3-5)^{2}}\)
= \(\sqrt{9+4}=\sqrt{13}\)
Now, BC + CA = \(\sqrt{13}+\sqrt{13}\) = 2√13 = AB
∴A, B and C are collinear then A, B, C and D do not form any quadrilateral.

(iii) Given points are A (4, 5); B (7, 6); C (4, 3) and D (1, 2)
AB = \(\sqrt{(7-4)^{2}+(6-5)^{2}}\)
= \(\sqrt{9+1}=\sqrt{10}\)

BC = \(\sqrt{(4-7)^{2}+(3-6)^{2}}\)
= \(\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\)

CD = \(\sqrt{(1-4)^{2}+(2-3)^{2}}\)
= \(\sqrt{9+1}=\sqrt{10}\)

DA = \(\sqrt{(4-1)^{2}+(5-2)^{2}}\)
= \(\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\)

AC = \(\sqrt{(4-4)^{2}+(3-5)^{2}}\)
= \(\sqrt{0+4}\) = 2

BD = \(\sqrt{(1-7)^{2}+(2-6)^{2}}\)
= \(\)

From above discussion, it is clear that AB = CD and BC = DA. and AC ≠ BD.
i.e., opposite sides are equal but their diagonals are not equal.
Hence, given quadrilateral ABCD is a parallelogram.

Question 7.
Find the points on the x-axis which is equidistant from (2, – 5) and (- 2, 9).
Solution:
Let required point be P (x, 0) and given points be A (2, – 5) and B (- 2, 9).
According to question,
PA = PB
(PA)² = (PB)²
or (2 – x)² + (- 5- 0)² = (- 2 – x)² + (9 – 0)²
or 4 + x² – 4x + 25 = 4 + x²+ 4x + 81
-8x = 56
x = \(\frac{4}{4}\) = – 7
Hence, required point be (- 7, 0).

Question 8.
Find the values of y for which the distance between the points P (2, – 3) and Q (10, y) is 10 units.
Solution:
Given points are P (2, – 3) and Q (10, y)
PQ = \(\sqrt{(10-2)^{2}+(y+3)^{2}}\)
= \(\sqrt{64+y^{2}+9+6 y}\)
= \(\sqrt{y^{2}+6 y+73}\)
According to question,
PQ = 10
or \(\sqrt{y^{2}+6 y+73}\) = 10
Squaring
or y² + 6y + 73 = 100
or y² + 6y – 27 = 0
or y² + 9y – 3y – 27 = 0
S = 6 P = – 27
or y (y + 9) – 3 (y + 9) = 0
or (y + 9) (y – 3) = 0
Either y + 9 = 0 or y – 3 = 0
y = – 9 or y = 3
Hence, y = – 9 and 3.

Question 9.
If Q (0, 1) is equidistant from P (5, – 3) and R (x, 6), find the values of x. Also find the distances QR and PR.
Solution:
Given points Q (0, 1); P (5, – 3) and R (x, 6)
QP = \(\sqrt{(5-0)^{2}+(-3-1)^{2}}\)
= \(\sqrt{25+16}=\sqrt{41}\)

and QR = \(\sqrt{(x-0)^{2}+(6-1)^{2}}\)
= \(\sqrt{x^{2}+25}\)

According to question,
QP = QR
or \(\sqrt{41}=\sqrt{x^{2}+25}\)
Squaring
or 41 = x² + 25
or x² = 16
or x = ± √16 = ± √4.

When x = 4 then R (4, 6).
QR = \(\sqrt{(4-0)^{2}+(6-1)^{2}}\)
= \(\sqrt{16+25}=\sqrt{41}\)

PR = \(\sqrt{(4-5)^{2}+(6+3)^{2}}\)
= \(\sqrt{1+81}=\sqrt{82}\)

When x = – 4 then R (- 4, 6).
QR = \(\sqrt{(-4-0)^{2}+(6-1)^{2}}\)
= \(\sqrt{16+25}=\sqrt{41}\)

PR = \(\sqrt{(-4-5)^{2}+(6+3)^{2}}\)
= \(\sqrt{81+81}=\sqrt{162}\).

Question 10.
Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4).
Solution:
Let required points be P (x, y) and given points are A (3, 6) and B (- 3, 4)
PA = \(\sqrt{(3-x)^{2}+(6-y)^{2}}\)
= \(\sqrt{9+x^{2}-6 x+36+y^{2}-12 y}\)
= \(\sqrt{x^{2}+y^{2}-6 x-12 y+45}\)

and PB = \(\sqrt{(-3-x)^{2}+(4-y)^{2}}\)
= \(\sqrt{9+x^{2}+6 x+16+y^{2}-8 y}\)
= \(\sqrt{x^{2}+y^{2}+6 x-8 y+25}\)

According to question,
PA = PB
\(\sqrt{x^{2}+y^{2}-6 x-12 y+45}\) = \(\sqrt{x^{2}+y^{2}+6 x-8 y+25}\)
sq,. both sides, we have,
or x² + y² – 6x – 12y + 45 = x² + y² + 6x – 8y – 25
or -12x – 4y + 20 = 0
or 3x + y – 5 = 0 is the required relation.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 5 Arithmetic Progressions Ex 5.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.1

Question 1.
In which of the following situations, does the list of numbers involved make an arithmetic progression, and why ?
(i) The taxi fare after each km when the fare is 15 for the first km and 8 for each additional km.

(ii) The amount of air present in a cylinder when a vacuum pump removes of the air remaining in the cylinder at a time.

(iii) The cost of digging a well after every metre of digging, when it costs 150 for the first metre and rises by 50 for each msubsequent metre.

(iv) The amount of money in the account every year when 10000 is deposited at compound interest at 8% per annum.

Solution:
(i) Let T_n denotes the taxi fare in n^th km.
According to question,
T₁ = 15 km;
T₂ = 15 + 8 = 23;
T₃ = 23 + 8 = 31
Now, T₃ – T₂ = 31 – 23 = 8
T₂ – T₁ = 23 – 15 = 8
Here, T₃ – T₂ = T₂ – T₁ = 8
∴ given situation form an AP.

(ii) Let T_n denotes amount of air present in a cylinder.
According to question,
T₁ = x;
T₂ = x – \(\frac{1}{4}\)x
= \(\frac{4-1}{4}\)x = \(\frac{3}{4}\)x
T₃ = \(\frac{3}{4} x-\frac{1}{4}\left[\frac{3}{4} x\right]=\frac{3}{4} x-\frac{3}{16} x\)

= \(\left(\frac{12-3}{16}\right) x=\frac{9}{16}\)x and so on
Now, T₃ – T₂ = \(\frac{9}{16}\)x – \(\frac{3}{4}\)x
= \(\left(\frac{9-12}{16}\right) x=-\frac{3}{16}\)x

T₂ – T₁ = \(\frac{3}{4}\)x – x
= \(\left(\frac{3-4}{4}\right) x=-\frac{1}{4}\)x
Here, T₃ – T₂ ≠ T₂ – T₁
∴ given situation donot form an AP.

(iii) Let T_n denotes cost of digging a well for the nth metre,
According to question,
T₁ = ₹ 150; T₂ = (150 + 50) = ₹ 200;
T₃ = ₹ (200 + 5o) = 250 and so on
Now, T₃ – T₂ = ₹ (250 – 200) = 50
T₂ – T₁ = ₹ (200 – 150) = 50
Here, T₃ – T₂ = T₂– T₁ = 50
∴ given situation form an A.P.

(iv) Let T_n denotes amount of money in the nth year.
According to question
T₁ = ₹ 10,000
T₂ = ₹ 10,000 + ₹ \(\left[\frac{10,000 \times 8 \times 1}{100}\right]\)
= ₹ 10,000 + ₹ 800 = ₹ 10,800
T₃ = ₹ 10,800 + ₹ \(\left[\frac{10,800 \times 8 \times 1}{100}\right]\)
= ₹ 10,800 + ₹ 864
= ₹ 11,640 and so on.
Now, T₃ – T₂ = ₹ (11,640 – 10,800) = ₹ 840
T₂ – T₁ = ₹ (10,800 – 10,000) = ₹ 800
Here, T₃ – T₂ ≠ T₂ – T₁
∴ given situation do not form an A.P.

Question 2.
Write first four terms of the AP, when the first term a and the common difference d are given as follows:
(1) a = 10, d = 10
(ii) a = -2, d = 0
(iii) a = 4, d = -3
(iv) a = -1, d = \(\frac{1}{2}\)
(w) a = -1.25, d = -0.25
Solution:
(i) Given that first term = a = 10
and common difference = d = 10
∴ T₁ = a = 10;
T₂ = a + d = 10 + 10 = 20;
T₃ = a + 2d
= 10 + 2 × 10 = 10 + 20 = 30;
T₄ = a + 3d = 10 + 3 × 10
= 10 + 30 = 40
Hence, first four terms of an A.P. are 10, 20, 30, 40………….

(ii) Given that first term = a = -2
and common iifference = d = 0
∴ T₁ = a = -2;
T₂ = a + d = -2 + 0 = -2
T₃ = a + 2d = -2 + 2 × 0 = -2
T₄ = a + 3d = -2 + 3 × 0 = -2
Hence, first four terms of an A.P. are -2, -2, -2, -2,…………….

(iii) Given that first term = a = 4
and common difference = d = -3
∴ T₁ = a = 4;
T₂= a + d = 4 – 3 = 1
T₃ = a + 2d = 4 + 2(-3) = 4 – 6 = -2
T₄ = a + 3d = 4 + 3(-3) = 4 – 9 = -5
Hence, first four terms of an A.P. are 4, 1, -2, -5, ……….

(iv) Given that first term = a = -1
and common difference = d = \(\frac{1}{2}\)
∴ T₁ = a = -1; T₂ = a + d
= -1 + \(\frac{1}{2}\) = \(-\frac{1}{2}\)
T₃ = a + 2d = -1 + 2(\(\frac{1}{2}\))
= -1 + 1 = 0
T₄ = a + 3d = -1 + 3(\(\frac{1}{2}\))
= \(\frac{-2+3}{2}=\frac{1}{2}\)
Hence, first four terms of an AP are -1, –\(\frac{1}{2}\), 0, \(\frac{1}{2}\), …………..

(v) Given that first term = a = – 1.25
and common difference = d = – 0.25
∴ T₁ = a = – 1.25;
T₂ = a + d = – 1.25 – 0.25 = -1.50
T₃ = a + 2d = – 1.25 + 2(- 0.25)
= – 1.25 – 0.50 = – 1.75
T₄ = a + 3d = – 1.25 + 3(- 0.25)
= – 1.25 – 0.75 = – 2
Hence, first four terms of an A.P. are – 1.25, – 1.50, – 1.75, – 2, ……………..

Question 3.
For the following APs, wilte the first term and the common difference:
(i) 3, 1, -1, -3, …………
(ii) 5, -1, 3, 7, ………….
(iii) \(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}\), …………..
(iv) 0.6, 1.7, 2.8, 3.9, ………..
Solution:
(i) Given A.P., is 3, 1, -1, -3, ………
Here T₁ = 3, T₂ = 1,
T₃ = -1, T₄ = -3
First term = T₁ = 3
Now, T₂ – T₁ = 1 – 3 = – 2
T₃ – T₂ = – 1 – 1 = -2
T₄ – T₃ = -3 + 1 = -2
∴ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = – 2
Hence, common difference = – 2 and first term = 3.

(ii) Given A.P. is – 5, – 1, 3, 7, ………….
Here T₁ = – 5, T₂ = – 1,
T₃ = 3, T₄ = 7
First term T₁ = -5
Now, T₂ – T₁ = -1 + 5 = 4
T₃– T₂ = 3 + 1 = 4
T₄ – T₃ = 7 – 3 = 4
∴ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = 4
Hence, common difference = 4 and first term = – 5.

(iii) Given AP. is:
\(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots\)
Here T₁ = \(\frac{1}{3}\), T₂ = \(\frac{5}{3}\),
T₃ = \(\frac{9}{3}\), T₄ = \(\frac{13}{3}\)
First term = T₁ = \(\frac{1}{3}\)
Now, T₂ – T₁ = \(\frac{5}{3}-\frac{1}{3}=\frac{5-1}{3}=\frac{4}{3}\)
T₃ – T₂ = \(\frac{9}{3}-\frac{5}{3}=\frac{9-5}{3}=\frac{4}{3}\)
T₄ – T₃ = \(\frac{13}{3}-\frac{9}{3}=\frac{13-9}{3}=\frac{4}{3}\)
∴ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = \(\frac{4}{3}\)

Hence, common difference = \(\frac{4}{3}\) and first term = \(\frac{1}{3}\).

(iv) Given A.P. is 0.6, 1.7, 2.8, 3.9,…
Here, T₁ = 0.6, T₂ = 1.7, T₃ = 2.8, T₄ = 3.9
First term = T₁ = 0.6
Now, T₂ – T₁ = 1.7 – 0.6 = 1.1
T₃ – T₂ = 2.8 – 1.7 = 1.1
T₄ – T₃ = 3.9 – 2.8 = 1.1
Hence, common difference = 1.1 and first term = 0.6.

Question 4.
WhIch of the following are APs? If they form an AP, find the common difference d and write three more terms.
(i) 2, 4, 8, 16
(ii) 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\), ………
(iii) – 1.2, – 3.2, – 5.2, – 7.2, ………….
(iv) – 10, – 6, – 2, 2, ………….
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2, ……….
(vi) 0.2, 0.22, 0.222, 0.2222, ………….
(vii) 0, -4, -8, -12, …………..
(viii) \(-\frac{1}{2}\), \(-\frac{1}{2}\), \(-\frac{1}{2}\), \(-\frac{1}{2}\), ………..
(ix) 1, 3, 9, 27 …………….
(x) a, 2a, 3a, 4a, ………………
(xi) a, a², a³, a⁴, ……………….
(xii) √2, √8, √18, √32, …………
(xiii) √3, √6, √9, √12, ……………..
(xiv) 1², 3², 5², 7², ………..
(xv) 1², 5², 7², 73, ………….

Solution:
(i) Given terms are 2, 4, 8, 16 ………………
Here T₁ = 2, T₂ = 4, T₃ = 8, T₄ = 16
T₂ – T₁ = 4 – 2 = 2
T₃ – T₂ = 8 – 4 = 4
∵ T₂ – T₁ ≠ T₃ – T₂
Hence, given terms do not form an A.P.

(ii) Given terms are 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\), ………
Here T₁ = 2, T₂ = 4, T₃ = 3, T₄ = 16
T₂ – T₁ = \(\frac{4}{4}\) – 2 = \(\frac{5-4}{2}\) = \(\frac{1}{2}\)
T₃ – T₂ = 3 – \(\frac{5}{2}\) = \(\frac{6-5}{2}=\frac{1}{2}\)
T₄ – T₃ = \(\frac{7}{2}-3=\frac{7-6}{2}=\frac{1}{2}\)
∵ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = \(\frac{1}{2}\)
∴ Common difference = d = \(\frac{1}{2}\)
Now, T₅ = a + 4d = 2 + 4\(\frac{1}{2}\) = 4

T₆ = a + 5d = 2 + 5(\(\frac{1}{2}\)) = \(\frac{4+5}{2}=\frac{9}{2}\)

T₇ = a + 6d = 2 + 6(\(\frac{1}{2}\)) = 2 + 3 = 5.

(iii) Given terms are – 1.2, – 3.2, – 5.2, – 7.2, …………
Here T₁ = – 1.2, T₂ = – 3.2,
T₃ = – 5.2, T₄ = – 7.2
T₂ – T₁ = – 3.2 + 1.2 = – 2
T₃ – T₂ = – 5.2 + 3.2 = – 2
T ₄ – T₃ = – 7.2 + 5.2 = – 2
∵ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = – 2
∴ Common difference = d = – 2
Now, T₅ = a + 4d = – 1.2 + 4(-2) = – 1.2 – 8 = – 9.2
T₆ = a + 5d = – 1.2 + 5(-2) = – 1.2 – 10 = – 11.2
T₇ = a + 6d = – 1.2 + 6(-2) = -1.2 – 12 = – 13.2

(iv) Given terms are – 10, – 6, – 2, 2, ………..
Here T₁ = – 10,T₂ = – 6
T₃ = – 2, T₄=2 .
T₂ – T₁ = – 6 + 10 = 4
T₃ – T₂ = – 2 + 6 =4
T₄ – T₃ = 2 + 2 = 4
∵ T₂ – T₁=T₃ – T₂ = T₄ – T₃ = 4 .
∴ Common difference = d = 4
Now, T₅ = a + 4d = – 10 + 4(4) = – 10 + 16 = 6
T₆ = a + 5d = – 10 + 5(4) = – 10 + 20 = 10
T₇ = a + 6d = – 10 + 6(4) = – 10 + 24 = 14.

(v) Given terms are 3, 3 + √2, 3 + 2√2, 3 + 3√2, …………
Here T₁ = 3, T₂ = 3 + √2,
T₃ = 3 + 2√2, T₄= 3 + 3√2
T₂ – T₁ = 3 + √2 – 3 = √2
T₃ – T2 = 3 + 2√2 – (3 + √2)
= 3 + 2√2 – 3 – √2 = √2
T₄ – T₃ = 3 + 3√2 – (3 + 2√2)
= 3 + 3√2 – 3 – 2√2 = √2
∵ T₂ -T₁ = T₃ – T₂ = T₄ – T₃ = √2
∴ Common difference = d = √2
Now, T₅ = a + 4d = 3 + 4(√2) = 3 + 4√2
T₆ = a + 5d = 3 + 5√2
T₇ = a + 6d = 3 + 6√2

(vi) Given terms are 0.2, 0.22, 0.222, 0.2222, …………..
Here Here T₁ = 0.2, T₂ = 0.22,
T₃ = 0.222, T₄ = 0.2222.
T₂ – T₁ = 0.22 – 0.2 = 0.02
T₃ – T₂ = 0.222 – 0.22 = 0.002
∵ T₂ – T₁ ≠ T₃ – T₂
∴ given terms do not form an A.P.

(vii) Given terms are 0, -4, -8, -12
Here T₁ = 0, T₂ = -4,
T₃ = -8, T₄ = -12
T₂ – T₁ = – 4 – 0 = -4
T₃ – T₂= – 8 + 4 = -4
T₄ – T₃= – 12 + 8 = -4.
T₂ – T₁ = T₃ – T₂ = T₄ – T₃
∴ Common difference = d = -4
Now, T₅= a + 4d = 0 + 4(-4) = -16
T₆ = a + 5d = 0 + 5(-4) = -20
T₇ = a + 6d = 0 + 6(-4) = -24.

(viii) Given terms are \(-\frac{1}{2}\), \(-\frac{1}{2}\), \(-\frac{1}{2}\), \(-\frac{1}{2}\), ……….
Here T₁ = \(-\frac{1}{2}\), T₂ = –\(\frac{1}{2}\)
T₃ = \(-\frac{1}{2}\), T₄ = \(-\frac{1}{2}\)
T₂ – T₁ = \(-\frac{1}{2}\) + \(\frac{1}{2}\) = 0
T₃ – T₂ = \(-\frac{1}{2}\) + \(\frac{1}{2}\) = 0
∵ T₂ – T₁ = T₃ – T₂ = 0
∴ Common difference = d = 0
Now, T₅ = T₆ = T₇ = –\(\frac{1}{2}\)
[∵ a = –\(\frac{1}{2}\), d = 0]

(ix) Given terms are 1, 3, 9, 27
T₁ = 1, T₂ = 3, T₃ = 9, T₄ = 27
T₂ – T₁ = 3 1 = 2
T₃ – T₂ = 9 – 3 = 6.
∵ T₂ – T₁ ≠ T₃ – T₂
∴ Given terms do not form an A.P.

(x) Given terms are a, 2a, 3a, 4a, …
T₁ = a, T₂ = 2a, T₃ = 3a, T4 = 4a
T₂ – T₁ = 2a – a = a
T₃ – T₂ = 3a – 2a = a
T₄ – T₃ = 4a – 3a = a
∵ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = a
∴ Common difference = d = a
Now T₅ = a + 4d = a + 4(a) = a + 4a = 5a
T₆ = a + 5d = a + 5a = 6a
T₇ = a + 6d = a + 6a = 7a

(xi) Given terms are a, a², a³, a⁴, …………
T₁ = a, T₂ = a², T3 = a³, T₄ = a⁴
T₂ – T₁ = a² – a
T₃ – T₂ = a³ – a²
∵ T₂ – T₁ ≠ T₃ – T₂
∴ Given terms do not form an A.P.

(xii) Given terms are √2, √8, √18, √32, …………
T₁ = √2, T₂ = √8, T₃ = √18, T₄ = √32
or T₁ = √2, T₂ = 2√2 T₃ = 3√2, T₄ = 4√2
T₂ – T₁ = 2√2 – √2 = √2
T₃ – T = 3√2 – 2√2 = √2
T₄ – T₃ = 4√2 – 3√2 = √2
∵ T₂ – T₁ = T₃ – T₂ = T₄ – T₃= √2
∴ Common difference = d = √2
Now, T₅ = a + 4d = √2 + 4√2 = 5√2
T₆ = a + 5d = √2 + 5√2 = 6√2
T₇ = a + 6d = √2 + 6√2 = 7√2

(xiii) Given terms are √3, √6, √9, √12, ……………..
T₁ = √3, T₂= √6, T₃= √9, T₄= √12
or T₁ = √3, T₂ = √6, T₃ = 3, T₄ = 2√3
T₄ – T₁ = √6 – √3
T₃ – T₂ = 3 – √6
∵ T₂ – T₁ ≠ T₃ – T₂
∴Given terms do not form an A.P.

(xiv) Given terms are 1², 3², 5², 7², ………..
T₁ = 1², T₂ = 3², T₃ = 5², T₄ = 7²
or T₁ = 1, T₂ = 9, T₃ = 25, T₄ = 49
T₄ – T₁ = 9 – 1 = 8
T₃ – T₂ = 25 – 9 = 16
∵ T₂ – T₁ ≠ T₃ – T₂
∴ Given terms do not form an A.P.

(xv) Given terms are 12, 52, 72, 73
T₁ = 12, T₂ = 52, T₃ = 72, T₄ = 73
or T₁ = 1, T₂ = 25, T₃ = 49, T₄ = 73
T₂ – T₁ = 25 – 1 = 24
T₃ – T₂ =49 – 24= 24
T₄ – T₃ = 73 – 49 = 24
∵ T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = 24
∴ Common difference = d = 24
T₅ = a + 4d = 1 + 4(24) = 1 + 96 = 97
T₆ = a + 5d = 1 + 5(24) = 1 + 120 = 121
T₇ = a + 6d = 1 +6(24) = 1 + 144 = 145

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 5 Arithmetic Progressions Ex 5.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.2

Question 1
Fill in the blanks in the following table, given that a is the first term, d the common difference and a the nth term of the
AP:

Solution:
(i) Here a = 7, d = 3, n = 8
∵ a_n = a + (n – 1)d
∴ a₈ = 7 + (8 – 1)3
= 7 + 21 = 28.

(ii) Here a = – 18, n = 10, a_n = 0
∵ a_n = a + (n – 1)d
∴ a₁₀ = – 18 + (10 – 1)d
or 0 = – 18 + 9d .
or 9d = 18
d = \(\frac{18}{2}\) = 2.

(iii) Here d = – 3, n = 18, a_n = – 5
∵ a_n = a + (n – 1)d
∴ a₁₈ = a + (18 – 1)(-3)
or -5 = a – 51
or a = – 5 + 51 = 46.

(iv) Here a = – 18.9, d = 2.5 a_n = 3.6
∵ a_n = a + (n – 1)d
∴ 3.6 = – 18.9 + (n – 1) 2.5
or 3.6 + 18.9 = (n – 1) 2.5
or (n – 1) 2.5 = 22.5
or n – 1 = \(\frac{22.5}{2.5}\)
or n = 9 + 1 = 10.

(v) Here a = 3.5, d = 0, n = 105
∵ a_n = a + (n – 1) d
∴ a_n = 3.5 + (105 – 1) 0
a_n = 3.5 + 0 = 3.5.

Question 2.
Choose the correct choice in the following and justify:
(i) 30th term of the AP: 10, 7, 4, …………….. is
(A) 97 (B) 77 (C) – 77 (D) – 87

(ii) 11th term of the AP: – 3, – \(\frac{1}{2}\), 2, ………. is
(A) 28 (B) 22 (C) – 38 (D) – 48\(\frac{1}{2}\)

Solution:
(i) Given A.P. is 10, 7, 4 ……………
T₁ = 10, T₂ = 7, T₃ = 4
T₂ – T₁ = 7 – 10 = – 3
T₃ – T₂ = 4 – 7 = – 3
∵ T₂ – T₁ = T₃ – T₂ = – 3 = d(say)
∵ T_n = a + (n – 1) d
Now, T₃₀ = 10 + (30 – 1)(-3)
= 10 – 87 = – 77
∴ Correct choice is (C).

(ii) Given A.P. is – 3, –\(\frac{1}{2}\), 2, ……….
T₁ = – 3 T₂ = –\(\frac{1}{2}\), T₃ = 2, …………..
T₂ – T₁ = –\(\frac{1}{2}\) + 3 = \(\frac{-1+6}{2}=\frac{5}{2}\)
T₃ – T₂ = 2 + \(\frac{1}{2}\) = \(\frac{4+1}{2}=\frac{5}{2}\)
∵ T₂ – T₁ = T₃ – T₂ = \(\frac{5}{2}\) = d(say)
∵ T_n = a + (n – 1) d
Now, T₁₁ = -3 + (11 – 1) \(\frac{5}{2}\)
= -3 + 10 × \(\frac{5}{2}\) = – 3 + 25 = 22
∴ Correct choice is (B).

Question 3.
In the following APs, find the missing terms in the boxes:
(i) 2, , 26
(ii), 13, , 3
(iii) 5,, , 9
(iv) – 4, , , , , 6
(v) , 38, , , , – 22
Solution:
Let a be the first term and ‘d’ be the common difference of given A.P.
(i) Here T₁ = a = 2
and T₃ = a + 2d = 26
or 2 + 2d = 26
or 2d = 26 – 2 = 24
or d = 12
∴ Missing term = T₂ = a + d = 2 + 12 = 14.

(ii) Here, T₂ = a + d = 13 ……………(1)
and T₄ = a + 3d = 3 …………….(3)
Now, (2) – (1) gives

Substitute this value of d in (1), we get
a – 5 = 13
a = 13 + 5 = 18.
∴ T₁ = a = 18
T₃ = a + 2d = 18 + 2(-5)
= 18 – 10 = 8.

(iii)Here T₁ = a = 5
and T₄ = a + 3d = 9
or a + 3d = \(\frac{19}{2}\)
or 5 + 3d = \(\frac{19}{2}\)
or 3d = \(\frac{19}{2}\) – 5
or 3d = \(\frac{19-10}{2}=\frac{9}{2}\)
or d = \(\frac{9}{2} \times \frac{1}{3}=\frac{3}{2}\)
T₂ = a + d = 5 + \(\frac{3}{2}\)
= \(\frac{10+3}{2}=\frac{13}{2}\)
T₃ = a + 2d = 5 + 2(\(\frac{3}{2}\)) = 5 + 3 = 8.

(iv) Here T₁ = a = —
T₆ = a + 5d = 6
or -4 + 5d = 6
or 5d = 6 + 4
or 5d = 10
or d = \(\frac{10}{2}\) = 2
Now, T₂ = a + d = -4 + 2 = -2
T₃ = a + 2d = – 4 + 2(2)
= – 4 + 4 = 0
T₄ = a + 3d = – 4 + 3(2)
= – 4 + 6 = 2
T₅ = a + 4d = – 4 + 4(2)
= – 4 + 8 = 4

(v) Here T₂ = a + d = 38 ………….(1)
and T₆ = a + 5d = -22 ……………(2)
Now, (2) – (1) gives

Substitute this value of d in (1), we get
a + (-15) = 38
a = 38 + 15 = 53
∴ T₁ = a = 53
T₃ = a + 2d = 53 + 2(-15) = 53 – 30 = 23.
T₄ = a + 3d = 53 + 3(-15) = 53 – 45 = 8
T₅ = a + 4d = 53 + 4(-15) = 53 – 60 = – 7.

Question 4
Which term of the A.P. 3, 8, 13, 18, …………… is 78?
Solution:
Given A.P. is 3, 8, 13, 18, ………….
T₁ = 3, T₂ = 8, T₃ = 13, T₄ = 18
T₂ – T₁ =8 – 3=5
T₃ – T₂= 13 – 8=5
T₂ – T₁ = T₃ – T,= 5 = d (say)
Using, T_n = a + (n – I) d
or 78 = 3 + (n – 1) 5
or 5(n – 1) = 78 – 3 = 75
or n – 1 = 15
or n = 15 + 1 = 16
Hence, 16th term of given AP. is 78.

Question 5.
Find the number of terms in each of the following APs:
(i) 7, 13, 19,…, 205
(ii) 18, 15\(\frac{1}{2}\), 13, ………….., – 47
Solution:
(i) Given A.P. is 7, 13, 19, …………..
T₁ = 7, T₂ = 13, T₃ = 19
T₂ – T₁ = 13 – 7 = 6
T₃ – T₂ = 19 – 13 = 6
T₂ – T₁ = T₃ – T₂ = 6 = d(say)
Using formula, T_n = a + (n – 1) d
205 = 7 + (n – 1) 6
or (n – 1) 6 = 205 – 7 = 198
or (n – 1) = \(\frac{198}{6}\)
or n – 1 = 33
n = 33 + 1 = 34
Hence, 34th term of an AP. is 205.

(ii) Given A P. is 18, 15\(\frac{1}{2}\), 13, …………..
T₁ = 18, T₂ = 15\(\frac{1}{2}\) = \(\frac{31}{2}\), T₃ = 13
T₂ – T₁ = \(\frac{31}{2}\) – 18 = \(\frac{31-36}{2}=-\frac{5}{2}\)
T₃ – T₂ = 13 – \(\frac{31}{2}\) = \(\frac{26-31}{2}=-\frac{5}{2}\)
∵ T₂ – T₁ = T₃ – T₂ = \(\frac{-5}{2}\) = d (say)
Using formula. T_n = a + (n – 1) d
– 47 = 18 + (n – 1) \(\frac{-5}{2}\)
or (n – 1) (\(\frac{-5}{2}\)) = – 47 – 18
or (n – 1) (\(\frac{-5}{2}\)) = – 65
or n – 1 = – 65 × – \(\frac{2}{5}\)
or n – 1 = 26
or n = 26 + 1 = 27
Hence, 27th term of an A.P. is – 47.

Question 6.
Is – 150 a term of 11, 8, 5, 2….? why?
Solution:
Given sequence is 11, 8, 5, 2, ………..
T₁ = 11, T₂ = 8, T₃ = 5, T₄ = 2
T₂ – T₁ = 8 – 11 = – 3
T₃ – T₂ = 5 – 8 = – 3
T₄ – T₃ = 2 – 5 = – 3
T₂ – T₁ = T₃ – T₂ = T₄ – T₃ = – 3 = d (say).
Let – 150 be any term of given A.P.
then T_n = – 150
a+(n – 1)d = – 150
or 11 +(n – 1)(- 3) = – 150
or (n – 1)( – 3) = – 150 – 11 = – 161
or n – 1 = \(\frac{161}{3}\)
or n = \(\frac{161}{3}\) + 1 = \(\frac{161+3}{3}\)
n = \(\frac{164}{3}\) = 54\(\frac{2}{3}\),
which is not a natural number.
Hence, – 150 cannot be a term of given A.P.

Question 7.
Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.
Solution:
Let ‘a’ and 4d’ be the first term and common difference of given A.P.
Given that T₁₁ = 38
a +(11 – 1) d = 38
[∵ T_n = a + (n – 1) d]
a + 10 d = 38
and T₁₆ = 73
a + (16 – 1) d = 73
[∵ T_n = a + (n – 1) d]
a + 15 d = 73
Now, (2) – (1) gives

Substitute this value of d in (1), we get
a + 10 (7) = 38
or a + 70 = 38
or a = 38 – 70 = – 32
Now, T₃₁ = a + (31 – 1) d
= – 32 + 30 (7) = – 32 + 210 = 178.

Question 8.
An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 291h term.
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of given A.P.
Given that, T₃ = 12
a + (3 – 1) d = 12
∵ T_n = a + (n – 1) d
or a + 2d = 12 ………………(1)
and Last term = T₅₀ = 106
a + (50 – 1) d = 106
∵ T_n = a + (n – 1) d
a + 49 d = 106 ……………(2)
Now, (2) – (1) gives

Substitute this value of d in (1), we get
a + 2(2) = 12
or a + 4 = 12
or a + 12 – 4 = 8
Now, T₂₉ = a + (29 – 1) d
= 8 + 28 (2) = 8 + 56 = 64.

Question 9.
If the 3rd and 9th tenus of an A.P. are 4 and – 8 respectively, which term of this A.P. is zero.
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of given AP.
Given that: T₃ = 4
a + (3 – 1) d = 4
∵ T_n = a + (n – 1) d
a + 2d = 4 …………..(1)
and T₉ = – 8
a + (9 – 1) d = 8
and T₉ = – 8
a + (9 – 1)d = 8
∵ T_n = a + (n – 1) d
or a + 8d = – 8
Now, (2) – (1) gives

Substitute this value of d in (1), we get
a + 2(- 2) = 4
or a – 4 = 4
or a = 4 + 4 = 8
Now, T_n = 0 (Given)
a + (n – 1) d = 0
or 8 + (n – 1)(- 2)=0
or -2 (n – 1) = – 8
or n – 1 = 4
or n = 4 + 1 = 5
Hence, 5th term of an AP. is zero.

Question 10.
The 17th term of an A.P. exceeds its 10th term by 7. Find the common difference.
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of given A.P.
Now, T₁₇ = a + (17 – 1) d = a + 16 d
and T₁₀ = a + (10 – 1) d = a + 9 d
According to question
T₁₇ – T₁₀ = 7
(a + 16 d) – (a + 9 d) = 7
or a + 16 d – a – 9 d = 7
7 d = 7
or d = \(\frac{7}{7}\) = 1
Hence, common difference is 1.

Question 11.
Which term of the A.P. 3, 15, 27, 39, …………. will be 132 more than its 54th term?
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of given A.P.
Given A.P. is 3, 15, 27, 39, …
T₁ = 3, T₂ = 15, T₃ = 27, T₄ = 39
T₂ – T₁ = 15 – 3 = 12
T₃ – T₂ = 27 – 15 = 12
:. d=T₂ – T₁ = T₃ – T₂ =12
Now, T₅₄ = a + (54 – 1) d
= 3 + 53 (12) = 3 + 636 = 639
According to question
T = T₅₄ + 132
a + (n – 1)d = 639 + 132
3 + (n – 1)(12) = 771
(n – 1) 12 = 771 – 3 = 768
or n – 1 = \(\frac{768}{12}\) = 64
or n = 64 + 1 = 65
Hence, 65th term of A.P. is 132 more than its 54th term.

Question 12.
Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of first AP.
Also, ‘A’ and ‘d’ be the first term and common difference of second A.P.
According to question
[T₁₀₀ of second A.P.] – [T₁₀₀ of first A.P.] = 100
or[A +(100 – 1)d] – [a +(100 – 1)d] = 100
or A + 99d – a – 99d = 100
or A – a = 100
Now, [T₁₀₀₀ of second A.P.] – [T₁₀₀₀ of first A.P.]
= [A + (1000 – 1) d) – (a + (1000 – 1) d]
= A + 999 d – a – 999 d
= A – a = 100 [Using (I)].

Question 13.
How many three-digits numbers are divisible by 7?
Solution:
Three digits numbers which divisible by 7 are 105, 112, 119 , 994
Here a = T₁ = 105, T₂ = 112, T₃ = 119 and T_n = 994
T₂ – T₁ = 112 – 105=7
T₃ – T₂ = 119 – 112=7
∴ d = T₂ – T₁ = T₃ – T₂ = 7
Given that, T_n = 994
a + (n – 1) d = 994
or 105 + (n – 1) 7 = 994
or (n – 1) 7 = 994 – 105
or (n – 1) 7 = 889
or n – 1 = \(\frac{889}{7}\) = 127
or n = 127 + 1 = 128.
Hence, 128 terms of three digit number are divisible by 7.

Question 14.
How many multiples of 411e between 10 and 250?
Solution:
Multiples of 4 lie between 10 and 250 are 12, 16, 20, 24, … 248
Here a = T₁ = 12, T₂ = 16, T₃ = 20 and T_n = 248
T₂ – T₁ = 16 – 12 = 4
T₃ – T₂ = 20 – 16 = 4
∴ d = T₂ – T₁ = T₃ – T₂ = 4
Given that, T_n = 248
a + (n – 1) d = 248
or 12 + (n – 1)4 = 248
or 4(n – 1) = 248 – 12 = 236
or n – 1 = \(\frac{236}{4}\) = 59
or n = 59 + 1 = 60
Hence, there are 60 terms which are multiples of 4 lies between 10 and 250.

Question 15.
For what value of n, are the n terms of two A.P.s 63, 65, 67, …………. and 3, 10, 17, …………….. equal?
Solution:
Given A.P. is 63, 65, 67, ……………..
Here a = T₁ = 63, T₂ = 65, T₃ = 67
T₂ – T1 = 65 – 63 = 2
T₃ – T₂ = 67 – 65 = 2
∴ d = T₂ – T₁ = T₃ – T₂ = 2
and second A.P. is 3, 10, 17, …
Here a = T₁ = 3, T₂ = 10, T₃ = 17
T₂ – T₁ = 10 – 3 = 7
T₃ – T₂ = 17 – 10 = 7
According to question.
[nth term of first A.P.] = [nth term of second A.P.]
63 + (n – 1)2 = 3 + (n – 1) 7
or 63 + 2n – 2 = 3 + 7n – 7
or 61 + 2n = 7n – 4
or 2n – 7n = – 4 – 61
– 5n = – 65
n = \(\frac{65}{5}\) = 13.

Question 16.
Determine the AP. whose third term is 16 and 7 term exceeds the by 12.
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of given A.P.
Given that T₃ = 16
a + (3 – 1) d = 16
a + 2d = 16
According to question
T₇ – T₅ = 12
[a + (7 – 1) d] – [a + (5 -1) d] = 12
a + 6 d – a – 44 = 12
2d = 12
d = \(\frac{12}{2}\) = 6
Substitute this value of d in (1), we get
a + 2(6) = 16
a = 16 – 12 = 4 .
Hence, given A.P. are 4, 10, 16, 22, 28, ………….

Question 17.
Find the 20th term from the last term of the AP: 3, 8, 13, ………., 253.
Solution:
Given A.P. is 3, 8, 13, …………., 253
Here, a = T₁ = 3, T₂ = 8, T₃ =13 and T_n = 253
T₂ – T₁ = 8 – 3 = 5
T₃ – T2 = 13 – 8 = 5
∴ d = T₂ – T₁ = T₃ – T₁ = 5
Now, T_n = 253
3 + (n – 1)5 = 253
∵ T_n = a + (n – 1) d
(n – 1) 5 = 250
n-1 = \(\frac{250}{5}\) = 50
n = 50 + 1 = 51
20th term from the end of AP = (Total number of terms) – 20 + 1
= 51 – 20 + 1 = 32nd term
∴ 20th term from the end of AP
= 32nd term from the starting
= 3 + (32 – 1)5
∵ Tn = a + (n – 1)d
= 3 + 31 × 5
= 3 + 155 = 158.

Question 18.
The sum of the 4th and 8th term of an AP is 24 and the sum of the 6’ and 10th terms is 44. FInd the first three terms of the A.P.
Solution:
Let ‘a’ and ‘d’ be the first term and common difference of given A.P.
According to 1st condition
T₄ + T₈ = 24
a + (4 – 1) d + a + (8 – 1) d = 24
∵ T_n = a + (n – 1) d
or 2a + 3d + 7d = 24
2a + 10d = 24
a + 5d = 12 …………(1)
According to 2nd condition
T₆ + T₁₀ = 44
a + (6 – 1) d + a +(10 – 1) d = 44
∵ T_n = a + (n – 1) d
2a + 5d + 9d = 44
2a + 14d = 44
a + 7d = 22
Now (2) – (1) gives

Substitute this value of d in (I). we get
a + 5(5) = 12
a + 25 = 12
a = 12 – 25 = -13
T₁ = a = -13
T₂ = a + d = 13 + 5 = -8
T₂ = a + 2d = – 13 + 2(5) = – 13 + 10 = -3
Hence, given A.P. is -13, -8, -3, ……………

Question 19.
Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000?
Solution:
Subba Rao’s starting salary = ₹ 5000
Annual increment = ₹ 200
Let ‘n’ denotes number of years.
∴ first term = a = ₹ 5000
Common diflerence = d = ₹ 200
and T_n = ₹ 7000
5000 + (n – 1) 200 = 7000
∵ T_n = a + (n – 1) d
(n – 1) 200 = 7000 – 5000
or (n – 1) 200 = 2000
or n – 1 = \(\frac{2000}{200}\) = 10
or n = 10 + 1 = 11
Now, in case of year the sequence are 1995. 1996, 1997, 1998, ……………
Here a = 1995, d = 1 and n = 11
Let T_n denotes the required year.
∴ T_n = 1995 + (11 – 1) 1
= 1995 + 10 = 2005
Hence, in 2005, Subba Rao’s salary becomes 7000.

Question 20.
Ramkali saved ₹ 5 in the first week of a year and then increased her weekly saving by ₹ 1.75. If in the nth week, her weekly saving becomes ₹ 20.75, find n.
Solution:
Amount saved in first week = ₹ 5
Increment in saving every week = ₹ 1.75
It is clear that, it form an A.P. whose terms are
T₁ = 5, d = 1.75
∴ T₂ = 5 + 1.75 = 6.75
T₃ = 6.75 + 1.75 = 8.50
Also. T_n = 20. 75 (Given)
5 + (n – 1) 1.75 = 20.75
∵ T_n = a + (n – 1) d
or (n – 1) 1.75 = 20.75 – 5
or (n – 1) 1.75 = 15.75
or (n – 1) = \(\frac{1575}{100} \times \frac{100}{175}\)
or n – 1 = 9
or n = 9 + 1 = 10
Hence, in 10th week, Ramkali’s saving becomes ₹ 20.75.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2

Question1.
A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of it.
Solution:
Radius of cone = Radius of hemisphere = 1 cm

R = 1 cm
Height of cone (H) = 1 cm
Volume of solid = volume of cone + volume of hemisphere
= \(\frac{1}{3}\) πR²H + \(\frac{2}{3}\) πR³
= \(\frac{1}{3}\) πR²2 [H + 2R]
= \(\frac{1}{3}\) π × 1 × 1 [1 + 2 × 1]
= \(\frac{1}{3}\) π × 3 = π cm³
Hence, Volume of solid = π cm³.

Question 2.
Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model the Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
Solution:

Radius of cone = Radius of cylinder (R) = \(\frac{3}{2}\) cm
∴ R = 1.5 cm
Height of eah cone (h) = 2 cm
∴ Height of cylinder = 12 – 2 – 2 = 8 cm
Volume of air in cylinder = volume of cylinder + 2 (volume of cone)

Volume of air in cylinder = \(\frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times \frac{28}{3}\)
= 22 × 3 = 66 cm³

Hence, Volume of air in cylinder =66 cm³.

Question 3.
A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm.

Solution:
Gulab Jamun is in the shape of cylinder
Diameter of cylinder = Diameter of hemisphere = 2.8 cm

Radius of cylinder = Radius of hemisphere (R)
= \(\frac{28.2}{2}\) = 1.4 cm
R = 1.4 cm
Height of cylindrical part = 5 – 1.4 – 1.4
= (5 – 2.8) cm = 2.2 cm.
Volume of one gulab Jamun = Volume of cylinder + 2 [Volume of hemisphere]
= πR²H + 2 \(\frac{2}{3}\) πR³

= πR² H + \(\frac{4}{3}\) R

= \(\frac{4}{4}\) × 1.4 × 1.4 2.2 + \(\frac{4}{3}\) × 1.4

= \(\frac{22}{7}\) × \(\frac{14}{10}\) × \(\frac{14}{10}\) 2.2 + \(\frac{4}{3}\) × 1.4

= \(\frac{22}{7}\) × \(\frac{196}{100}\) [2.2 + 1.86]

= \(\frac{22 \times 28}{100}\) [4.06]

Volume of one gulab Jamun = 25.05 cm³
Now volume of 45 gulab Jamuns = 45 × 25.05 cm³ = 1127.25 cm³
Volume of sugar syrup = 30% volume of 45 gulab Jamuns
= \(\frac{30 \times 1127.25}{100}\) = 338.175 cm3
Hence, Approximately sugar syrup = 338 cm³.

Question 4.
A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.

Solution. Length of cuboid (L) = 15 cm
Width of cuboid (B) = 10 cm
Height of cuboid (H) = 3.5 cm
Radius of conical cavity (r) = 0.5 cm
Height of conical cavity (h) = 1.4 cm
Volume of wood in Pen stand = volume of cuboid – 4 [volume of cone]
= LBH – 4 \(\frac{1}{3}\) πr²h

= 15 × 10 × 3.5 – \(\frac{4}{3}\) × \(\frac{22}{7}\) × 0.5 × 0.5 × 1.4

= \(\frac{15 \times 10 \times 35}{10}-\frac{4}{3} \times \frac{22}{7} \times \frac{5}{10} \times \frac{5}{10} \times \frac{14}{10}\)

= \(15 \times 35-\frac{22}{3 \times 5}\)
= 525 – 1.466 = 523.534 cm³
Hence, Volume of wood in Pen stand = 523.53 cm³.

Question 5.
A vessel is in the form of an inverted cone. Its height ¡s 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Solution:

Radius of cone (R) = 5 cm
Height of cone (H) = 8 cm
Radius of each spherical lead shot (r) = 0.5 cm
Let number of shot put into the cone = N
According to Question,
N [Volume of one lead shot] = \(\frac{1}{4}\) Volume of water in cone

= 10 × 10 = 100
Hence, Number of lead shots = 100.

Question 6.
A solid iron pole consists of a cylinder of height 220 cfi and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of Iron has approximately 8 g mass. (Use n = 3.14)
Solution:
Diameter of lower cylinder = 24 cm
Radius of lower cylinder (R) = 12 cm
Height of lower cylinder (H) = 220 cm
Radius of upper cylinder (r) = 8 cm
Height of upper cylinder (h) = 60 cm
Volume of pole = Volume of Lower cylinder + volume of upper cylinder
= πR²H + πr²h
= 3.14 × 12 × 12 × 220 + 3.14 × 8 × 8 × 60
= 99475.2 + 12057.6

Volume of pole = 111532.8 cm³
Mass of 1 cm³ = 8 gm
Mass of 111532.8 cm³ = 8 × 111532.8 = 892262.4 gm
= \(\frac{892262.4}{1000}\) kg = 892.2624 kg
Hence, Mass of Pole = 892.2624 kg.

Question 7.
A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
Solution:
Radius of cone = Radius of hemisphere = Radius of cylinder

Height of cone (h) = 120 cm
Height of cylinder (H) = 180 cm
Volume of cylindrical vessel = πR²H
= \(\frac{22}{7}\) × 60 × 60 × 180 = 2036571.4 cm³
Volume of solid inserted in cylinder = Volume of hemisphere + Volume of cone
= \(\frac{2}{3}\) πR³ + \(\frac{1}{3}\) πR²h

= \(\frac{1}{3}\) πR² [2R + h]

= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 60 × 60 [2 × 60 + 120]

= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 3600 [120 + 120]

= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 3600 × 240 = 905142.86 cm³
Volume of water flows out = 90514186 cm³
∴ Volume of water left in cylinder = Volume of cylinder – Volume of solid inserted in th’e vessel
= (2036571.4 – 905142.86) cm³ = 1131428.5 cm³
= \(\frac{1131428.5}{100 \times 100 \times 100}\) m³ = 1.131 m³
Hence, Volume of water left in cylinder = 1.131 m³.

Question 8.
A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.
Solution:
Diameter of neck (cylindrical Portion) = 2 cm

Radius of neck (r) = 1 cm
Height of cylindrical portion (H) = 8 cm
Diameter of spherical portion = 8.5 cm
Radius of spherical portion (R) = \(\frac{8.5}{2}\) cm = 4.25 cm
Volume of water in vessel = Volume of sphere + Volume of cylinder
= \(\frac{4}{3}\) πR³ + πR²h
= \(\frac{4}{3}\) × 3.14 × 4.25 × 4.25 × 4.25 × 3.14 × 1 × 1 × 8
= 321.39 + 25.12 = 346.51 cm³
Hence, Volume of water in vessel = 346.51 cm³ and She is wrong.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1

Question 1.
2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Solution:
Let side of cube = x cm
Volume of cube = 64 cm³
[volume of cube = (side)³]
x³ = 64
x = \(\sqrt[3]{4 \times 4 \times 4}\)
x = 4 cm
∴ side of cube = 4 cm.

When cubes are joined end to end and cuboid is formed
whose Length = 2x cm = 2(4) = 8 cm
Width = x cm = 4 cm
Height = x cm = 4 cm

Surface area of cuboid = 2[LB + Bh + hL]
= 2 [8 × 4 + 4 × 4 + 4 × 8]
= 2 [32 + 16 + 32]
= 2 [80]
∴ Surface area of cuboid = 160 cm².

Question 2.
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere ¡s 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Solution:

Diameter of hemisphere = Diameter of cylinder
= 14 cm
2R = 14 cm
Radius of hemisphere (R) = 7 cm
Total height of vessel = 13 cm
∴ Height of cylinder = (13 – 7) = 6 cm
Inner surface area of vessel = inner surface area of cylinder + Inner surface area of hemisphere
= 2πRH + 2πR²
= 2πR [H + R]
= 2 × \(\frac{22}{7}\) × 7(16 + 7)
= 44 × 13 = 572 cm²
Hence, Inner surface area of vessel = 572 cm²

Question 3.
A toy is ¡n the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Solution:
Radius of cone = Radius of hemisphere (R) = 3.5 cm
Total height of toy = 15.5 cm
∴ Height of cone (H) = (15.5 – 3.5) = 12 cm

Slant height of cone = \(\sqrt{\mathrm{R}^{2}+\mathrm{H}^{2}}\)

= \(\sqrt{(3.5)^{2}+(12)^{2}}\)

= \(\sqrt{12.25+144}\) = \(\sqrt{156.25}\)
Slain height conk (l) = 12.5 cm
Total surface area of toy = Surface area of cone + Surface area of hemisphere
= πRL + 2πR²
= πR[L + 2R]
= \(\frac{22}{7}\) × 3.5 [12.5 + 2 (3.5)1 cm²
= \(\frac{22}{7}\) × 3.5 [19.5] cm²
= \(\frac{1501.5}{7}\) = 214.5 cm²
∴ Total surface area of toy = 214.5 cm²

Question 4.
A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have?
Find the surface area of the solid.
Solution:
Side of cubical box = 7 cm
Diameter of hemisphere = Side of cubical box = 7 cm
2R = 7
R = \(\frac{7}{2}\) cm

Surface area of solid = (surface area of the cube) – (area of base of hemisphere) + curved surface area of hemisphere)
= 6l² – πR² + 2πR²
= 6l² + πR²
= 6(7)² + \(\frac{22}{7}\) \(\frac{7}{2}\)²
= [6 × 49 + 11 × \(\frac{7}{2}\)]cm²
= 294 + 38.5 = 332.5 cm²

Question 5.
A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of (he hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Solution:
Let each side of cube = a
∴ Diameter of hemisphere = Side of cube
2R = a
R = \(\frac{a}{2}\)

Surface area of remaining solid = Total surface area of euboid – Area of the top of cube + Inner curved Surface area of hemisphere
= 6 (side)² – πR² + 2πR²
= 6(a)² + πR²
= 6(a)² + π \(\frac{a}{2}\)²
= 6a² + π \(\frac{a^{2}}{4}\)
= a² 6 + \(\frac{\pi}{4}\) cm²

Question 6.
A medicine capsule is ¡n the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Solution:
Diameter of capsule = Diameter of hemisphere = Diameter of cylinder = 5 mm
∴ 2R = 5 mm
R = \(\frac{5}{2}\) mm

Length of entire capsule = 14 mm
Height of cylinderical part = (14 – \(\frac{5}{2}\) – \(\frac{5}{2}\)) mm
= (14 – 5) mm
H = 9 mm
Surface area of capsule = Surface area of cylinder + 2 Surface area of hemisphere
= 2πRH + 2 (2πR²)
= 2πRH + 4πR²
= 2πR [H + 2R]
= 2 × \(\frac{22}{7} \times \frac{5}{2}\left[9+2\left(\frac{5}{2}\right)\right]\)
= 2 × \(\frac{22}{7}\) × \(\frac{5}{2}\) [9 + 5]
= \(\frac{22}{7}\) × 5 × 14
= 220 mm²
Hence, Surface area of capsule = 220 mm².

Question 7.
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of 500 per (Note that the base of the tent will not be
covered with canvas.)
Solution:
Diameter of cone = Diameter of cylinder
2R = 4
R = 2 m

Radius of cone = Radius of cylinder
Height of cylinder (H) = 2.1 in
Slant height of cone (L) = 2.8 m
Curved surface area of tent = Curved surface of cylinder + Curved surface of conical part
= 2πRH + πRL
= πR [2H + L]
= \(\frac{22}{7}\) × 2[2(2.1) + 2.8]
= \(\frac{22}{7}\) × 2[4.2 + 2.8]
= \(\frac{22}{7}\) × 2 × 7
= 44 m²
∴ Curved surface area of tent = 44 m²
Cost of 1m² canvas = ₹ 500
Cost of 44 m² canvas = 44 × 500 = ₹ 22000
Hence, Total cost of canvas = ₹ 22000.

Question 8.
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.
Solution:
Diameter of cylinder (D) = 1.4 cm = Diameter of cone
∴ Radius of cylinder = Radius of cone (R) = 0.7 cm
Height of cylinder (H) = 2.4 cm

As we know, L² = R² + H² + (2.4)²
L = \(\sqrt{(0.7)^{2}+(2.4)^{2}}\)
= \(\sqrt{0.49+5.76}\) = \(\sqrt{6.25}\)
L = 2.5 cm
Total surface area of remaining solid = curved surface area of cylinder + Area of base of cylinder + Surface area of cone
= 2πRH + πR² + πRL
= πR [2R +R + L]
= \(\frac{22}{7}\) × 0.7 [2(2.4) + 0.7 + 2.5]

=\(\frac{22}{7}\) × \(\frac{7}{10}\) [4.8 + 3.2]

= \(\frac{22}{10}\) [8]

= \(\frac{176}{10}\) = 17.6 cm²

Hence, Total surface area remaining solid to nearest cm² = 18 cm².

Question 9.
A wooden article was made by scooping out a hemisphere form each end of a solid cylinder, as shown in Fig.

If the height of the cylinder is 10 cm, and its base ¡s of radius 3.5 cm, find the total surface area of the article.
Solution:
Height of cylinder (H) = 10 cm
Radius of cylinder = Radius of hemisphere (R) = 3.5 cm

Surface area of article = curved surface area of cylinder + 2 curved surface area of hemisphere
= 2πRH + 2 (2πR²)
= 2πR [H + 2R]
= 2 × \(\frac{22}{7}\) × 3.5 [10 + 2(3.5)]
= \(\frac{44}{7}\) × \(\frac{35}{10}\) [10 + 7]
= 44 × \(\frac{5}{10}\) × 17
= 44 × \(\frac{1}{2}\) × 17
= 22 × 17 = 374 cm²
Hence, total surface area of article = 374 cm².