Class 11

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 9 Sequences and Series Ex 9.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1

Question 1.
Write the first five terms of the sequence whose nth term is a_n = n (n + 2).
Answer.
a_n = n(n +2)
Substituting n = 1, 2, 3, 4 and 5, we obtain
a₁ = 1 (1 + 2) = 3,
a₂ = 2 (2 + 2) = 8,
a₃ = 3 (3 + 2) = 15,
a₄ = 4 (4 + 2) = 24,
a₅ = 5 (5 + 2) = 35
Therefore, the required terms are 3, 8, 15, 24 and 35.

Question 2.
Write the first five terms of the sequence whose nth term is a_n = \(\frac{n}{n+1}\).
Answer.
a_n = \(\frac{n}{n+1}\)
Sustituting n = 1, 2, 3, 4, 5, we otain
a_n = \(\frac{1}{1+1}=\frac{1}{2}\)

a_n = \(\frac{2}{2+1}=\frac{2}{3}\)

a_n = \(\frac{3}{3+1}=\frac{3}{4}\)

a_n = \(\frac{4}{4+1}=\frac{4}{5}\)

a_n = \(\frac{5}{5+1}=\frac{5}{6}\)

Therefore, the required terms are \(\frac{1}{2}\), \(\frac{2}{3}\) , \(\frac{3}{4}\) , \(\frac{4}{5}\) and \(\frac{5}{6}\) .

Question 3.
Write the first five terms of the sequence whose nth term is a_n = 2ⁿ.
Answer.
a_n = 2ⁿ
Substituting n = 1, 2, 3, 4, 5, we obtain
a₁ = 2¹ = 2
a₂ = 2² = 4
a₃ = 2³ = 8
a₄ = 2⁴ = 16
a₅ = 2⁵ = 32
Therefore, the required terms are 2, 4, 8, 16 and 32.

Question 4.
Write the first five terms of the sequence whose nth term is a_n = \(\frac{2 n-3}{6}\).
Answer.
Substituting n = 1,2, 3, 4, 5, we obtain

Question 5.
WrIte the first five terms of the sequence whose nth term is
a_n = (- 1)^{n – 1} 5^{n + 1}.
Ans.
Substituting n = 1,2, 3, 4, 5, we obtain
a₁ = (- 1)^{1 – 1} 5^{1 + 1} = 5² = 25,
a₂ = (- 1)^{2 – 1} 5^{2 + 1} = – 5³ = -125,
a₃ = (- 1)^{3 – 1} 5^{3 + 1} = 5⁴ = 625,
a₄ = (- 1)^{4 – 1} 5^{4 + 1} = 5⁵ = – 3125,
a₅ = (- 1)^{5 – 1} 5^{5 + 1} = 5⁶ = 15625
Therefore, the required terms are 25, – 125, 625, – 3125, and 15625.

Question 6.
Write the first five terms of the sequence whose nth term is a_n = \(n \frac{n^{2}+5}{4}\).
Answer.
Substituting n = 1, 2, 3, 4, 5, we obtain

Question 7.
Find the indicated terms in the following sequence whose nth term is a_n =4n – 3; a₁₇, a₂₄.
Answer.
We have a_n = 4n -3
On putting n =17, we get
a₁₇ = 4 × 17 – 3
= 68 – 3 = 65
On putting n = 24, we get
a₂₄ = 4 × 24 – 3
= 96 – 3 = 93.

Question 8.
Find the indicate term in the following sequence whose nth term is a_n = \(\frac{n^{2}}{2^{n}}\); a₇.
Answer.
Substituting n = 7, we obtain
a₇ = \(\frac{7^{2}}{2^{7}}=\frac{49}{128}\).

Question 9.
Find the indicated term in the following sequence whose nt1 term is a_n = (- 1)^{n – 1} n³; a₉.
Answer.
Substituting n = 9, we obtain
a₉ = (- 1)^{9 – 1} (9)³ = 729.

Question 10.
Find the indicated term in the following sequence whose nth term is a_n = \(\frac{n(n-2)}{n+3}\); a₂₀.
Answer.
Substituting n = 20, we obtain
a₂₀ = \(\frac{20(20-2)}{20+3}=\frac{20(18)}{23}=\frac{360}{23}\).

Question 11.
Write the first five terms of the following sequence and obtain the corresponding series:
a₁ = 3, a_n = 3a_{n – 1} + 2 for all n > 1.
Answer.
a₁ = 3, a_n = 3 a_{n – 1} + 2 for all n > 1
=> a₂ = 3a_{2 – 1} + 2
= 3a₁ + 2
= 3(3) + 2 = 11

a₃ = 3a_{3 – 1} + 2
= 3a₂ + 2
= 3(11) + 2 = 35

a₄ = 3a_{4 – 1} + 2
= 3a₃ + 2
= 3(35) + 2 = 107

a₅ = 3a_{5 – 1} + 2
= 3a₄ + 2
= 3(107) + 2 = 323.
Hence, the first five terms of the sequence are 3, 11, 35, 107 and 323.
The corresponding series is 3 + 11 + 35 + 107 + 323 + …

Question 12.
Write the first five terms of the following sequence and obtain the corresponding series:
a₁ = – 1, a_n = \(\frac{\boldsymbol{a}_{n-1}}{n}\), n > 2
Answer.

Question 13.
Write the first five terms of the following sequence and obtain the corresponding series:
a₁ = a₂ = 2, a_n = a_{n – 1} – 1, n > 2.
Answer.
a₁ = a₂ = 2,
a_n = a_{n – 1}, n > 2
=> a₃ = a₂ – 1 = 2 – 1 = 1,
a₄ = a₃ – 1 = 1 – 1 = 0
a₅ = a₄ – 1 = 0 – 1 = – 1.
Hence, the first five terms of the sequence are 2, 2, 1, 0 and – 1.
The corresponding series is 2 + 2 +1 + 0+ (- 1) + ……….

Question 14.
The Fibonacci sequence is defined by 1 = a₁ = a₂ and a_n = a_{n – 1} + a_{n – 2} n > 2. Find \(\frac{\boldsymbol{a}_{n+1}}{\boldsymbol{a}_{n}}\), for n = 1, 2, 3, 4, 5.
Answer.
1 = a₁ = a₂
a_n = a_{n – 1} + a_{n – 2}, n > 2
a₃ = a₂ + a₁
= 1 + 1 = 2,
a₄ = a₃ + a₂
= 2 + 1 = 3,
a₅ = a₄ + a₃
= 3 + 2 = 5,
a₆ = a₅ + a₃
= 5 + 3 = 8

For n = 1,
\(\frac{a_{n+1}}{a_{n}}=\frac{a_{2}}{a_{1}}=\frac{1}{1}\) = 1

For n = 2,
\(\frac{a_{n+1}}{a_{n}}=\frac{a_{3}}{a_{2}}=\frac{2}{1}\) = 2

For n = 3,
\(\frac{a_{n+1}}{a_{n}}=\frac{a_{4}}{a_{3}}=\frac{3}{2}\)

For n = 4,
\(\frac{a_{n+1}}{a_{n}}=\frac{a_{5}}{a_{4}}=\frac{5}{3}\)

For n = 5,
\(\frac{a_{n+1}}{a_{n}}=\frac{a_{6}}{a_{5}}=\frac{8}{5}\)

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 8 Binomial Theorem Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 8 Binomial Theorem Miscellaneous Exercise

Question 1.
Find a, b and n in the expansion of (a + b)ⁿ if the first three terms of the expansion are 729, 7290 and 30375, respectively.
Answer.
It is known that (r + 1)th term, (T_{r + 1}), in the binomial expansion of (a + b)ⁿ is given by T_{r + 1} = \({ }^{n} C_{r}\) a^{n – r r= a}
The first three terms of the expansion are given as 729, 7290, and 30375 respectively.
Therefore, we obtain
T₁ = \({ }^{2} C_{0} a^{n-0} b^{0}\)
= aⁿ = 729 ……………..(i)

T₂ = \({ }^{n} C_{1} a^{n-1} b^{1}\)
= n a^{n – 1} b = 7290 …………….(ii)

T₃ = \({ }^{n} C_{2} a^{n-2} b^{2}\)
= \(\frac{n(n-1)}{2} a^{n-2} b^{2}\) = 30375 ……………….(iii)

Dividing eq. (ii) by eq. (i), we obtain \(\frac{n a^{n-1} b}{a^{n}}=\frac{7290}{729}\)
⇒ \(\frac{n b}{a}\) = 10 ……………….(iv)

Dividing eq. (iii) by eq. (ii), we obtain

Question 2.
Find a if the coefficient of x² and x³ in the expansion of (3 + ax)⁹ are equal.
Answer.
It is known that (r + 1)th term, (T_{r + 1}), in the binomial expansion of (a + b)ⁿ is given by T_{r + 1} = \({ }^{n} C_{r} a^{n-r} b^{r}\)

Assuming that x² occurs in the (r + 1)th term in the expansion of (3 + ax)⁹, we obtain
T_{r + 1} = \({ }^{9} C_{r}(3)^{9-r}(a x)^{r}\)
= \({ }^{9} C_{r}(3){ }^{9-r} a^{r} x^{r}\)

Comparing the indices of x in x² and in T_{r + 1} we obtain r = 2
Thus, the coefficient of x² is
\({ }^{9} C_{2}(3)^{9-2} a^{2}=\frac{9 !}{2 ! 7 !}(3)^{7} a^{2}=36(3)^{7} a^{2}\)

Assuming that x³ occurs in the (k + 1)th term in the expansion of (3 + ax)⁹, we obtain
T_{k + 1} = \({ }^{9} C_{k}(3)^{9-k}(a x)^{k}={ }^{9} C_{k}(3)^{9-k} a^{k} x^{k}\)

Comparing the indices of x in x³ in T_{r + 1}, we obtain k = 3
Thus, the coefficient of x³ is
\({ }^{9} C_{3}(3)^{9-3} a^{3}=\frac{9 !}{3 ! 6 !}(3)^{6} a^{3}=84(3)^{6} a^{3}\)

It is given that the coefficients of x² and x³ are same.
84 (3)⁶ a³ = 36 (3)⁷ a²
⇒ 84a = 36 × 3
⇒ a = \(\frac{36 \times 3}{84}=\frac{108}{84}\)
⇒ a = \(\frac{9}{7}\)

Thus, the required value of a is \(\frac{9}{7}\).

Question 3.
Find the coefficient of x5 in the product (1 + 2x)⁶(1 – x)⁷ using binomial theorem.
Answer.
Using Binomial Theorem, the expressions, (1 + 2x)⁶ and (1 – x)⁷, can be expanded as

The complete multiplication of the two brackets is not required to be carried out.
Only those terms, which involve x⁵, are required.
The terms containing x⁵ are
1 (- 21x⁵) + (12x) (35x⁴) + (60x²) (- 35x³) + (160x³) (21x²) + (240x⁴) (- 7x) + (192 x⁵)(1)
= 171 x⁵
Thus, the coefficient of x⁵ in the given product.
= – 21 + 420 – 2100 + 3360 -1 680 +192 = 171.

Question 4.
If a and b are distinct integers, prove that a – b is a factor of aⁿ – bⁿ, whenever n is a positive integer.
[Hint : write aⁿ = (a – b + b)ⁿ and expand]
Answer.
In order to prove that (a – b) is a factor of (aⁿ – bⁿ), it has to be proved that
aⁿ – bⁿ = k(a – b), where k is some natural number
It can be written that, a = a – b + b
∴ aⁿ = (a – b + b)ⁿ = [(a – b) + b]ⁿ
= \({ }^{n} C_{0}(a-b)^{n}+{ }^{n} C_{1}(a-b){ }^{n-1} b+\ldots+{ }^{n} C_{n-1}(a-b) b^{n-1}+{ }^{n} C_{n} b^{n}\)

= \((a-b)^{n}+{ }^{n} C_{1}(a-b)^{n-1} b+\ldots+{ }^{n} C_{n-1}(a-b) b^{n-1}+b^{n}\)

\(a^{n}-b^{n}=(a-b)\left[(a-b)^{n-1}+{ }^{n} C_{1}(a-b)^{n-2} b+\ldots+{ }^{n} C_{n-1} b^{n-1}\right]\)

aⁿ – bⁿ = k (a -b)

where k = \(\left[(a-b)^{n-1}+{ }^{n} C_{1}(a-b)^{n-2} b+\ldots+{ }^{n} C_{n-1} b^{n-1}\right]\) is a natural number
This shows that (a – b) is a factor of (aⁿ – bⁿ), where n is a positive integer.

Question 5.
Evaluate (√3 + √2)⁶ – (√3 – √2)⁶.
Answer.
Firstly, the expression (a + b)⁶ – (a – b)⁶ is simplified by using Binomial Theorem.
This can be done as
(a + b)⁶ = \({ }^{6} C_{0}\) a⁶ + \({ }^{6} C_{1}\) a⁵ b + \({ }^{6} C_{2}\) a⁴ b² + \({ }^{6} C_{3}\) a³b³ + \({ }^{6} C_{4}\) a²b⁴ + \({ }^{6} C_{5}\) ab⁵ + \({ }^{6} C_{6}\) b⁶

= a⁶ + 6 a⁵b + 15 a⁴b² + 20 a³b³ + 15 a²b⁴ + 6 ab⁵ + b⁶ (a – b)⁶

= \({ }^{6} C_{0}\) a⁶ – \({ }^{6} C_{1}\) a⁵b + \({ }^{6} C_{2}\) a⁴b² – \({ }^{6} C_{3}\) a³b³ + \({ }^{6} C_{4}\) a²b⁴ – \({ }^{6} C_{5}\) a¹b⁵ + \({ }^{6} C_{6}\) b⁶

= a⁶ – 6 a⁵b + 15 a⁴b² – 20 a³b³ + 15 a²b⁴ – 6ab⁵ + b⁶

.-. (a + b)⁶ – (a – b)⁶ = 2[6a⁵b + 20a³b³ + 6ab⁵]

Putting a = √3 and b = √2, we obtain (√3 + √2)⁶ – (√3 – √2)⁶
= 2[6(√3)⁵ (√2) + 20 (√3)³ (√2)³ + 6 (√3) (√2)⁵]
= 2[54 √6 +120 √6 + 24 √6]
= 2 × 198 √6 = 396√6.

Question 6.
Find the value of (a² + \(\sqrt{a^{2}-1}\))⁴ + (a² – \(\sqrt{a^{2}-1}\))⁴.
Answer.
Firstly, the expression (x + y)⁴ + (x – y)⁴ is simplified by using Binomial Theorem.
This can be done as
(x + y)⁴ = \({ }^{4} C_{0}\) x⁴ + \({ }^{4} C_{1}\) x³y + \({ }^{4} C_{2}\) x²y² + \({ }^{4} C_{3}\) xy³ + \({ }^{4} C_{4}\) y⁴
= x⁴ + 4x³y + 6x²y² + 4xy³ + y⁴

(x – y)⁴ = \({ }^{4} C_{0}\) x⁴ – \({ }^{4} C_{1}\) x³y + \({ }^{4} C_{2}\) x²y² – \({ }^{4} C_{3}\) xy³ + \({ }^{4} C_{4}\) y⁴
= x⁴ – 4x³y + 6x²y² – 4xy³ + y⁴

∴ (x + y)⁴ + (x – y)⁴ = 2(x⁴ + 6x²y² + y⁴).

Putting x = a² and y = \(\sqrt{a^{2}-1}\), we obtain
(a² + \(\sqrt{a^{2}-1}\))⁴ + (a² – \(\sqrt{a^{2}-1}\))⁴
= 2 [(a²)⁴ + 6(a²)² (Ja2 -l)² + \(\left(\sqrt{a^{2}-1}\right)\)⁴]
= 2 [a⁸ + 6a⁴ (a² – 1) + (a² – 1)²]
= 2 [a⁸ + 6a⁶ – 6a⁴ + a⁴ – 2a² + 1]
= 2 [a⁸ + 6a⁶ – 5a⁴ – 2a² + 1]
= 2a⁸ + 12a⁶ – 10a⁴ – 4a² + 2.

Question 7.
Find an approximation of (0.99)⁵ using the first three terms of its expansion.
Answer.
We have, (0.99)⁵ = (1 – 0.01)⁵
= 1 – \({ }^{5} C_{1}\) × (0.01) + \({ }^{5} C_{2}\) × (0.01)² – ……….
= 1 – 0.05 + 10 × 0.0001 – ………..
= 1.001 – 0.05 = 0.951.

Question 8.
Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expression of \(\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^{n}\) is √6 : 1.
Answer.

2n – 16 = 4
⇒ 2n = 20
⇒ n = \(\frac{20}{2}\)
⇒ n = 10.

Question 9.
Expand using Binomial Theorem \(\left(1+\frac{x}{2}-\frac{2}{x}\right)^{4}\), x ≠ 0.
Answer.
Using Binomial Theorem, the given expression \(\left(1+\frac{x}{2}-\frac{2}{x}\right)^{4}\) can be expanded as \(\left[\left(1+\frac{x}{2}\right)-\frac{2}{x}\right]^{4}\),

= \(1+2 x+\frac{3}{2} x^{2}+\frac{x^{3}}{2}+\frac{x^{4}}{16}-\frac{8}{x}-12-6 x\) – \(x^{2}+\frac{8}{x^{2}}+\frac{24}{x}+6-\frac{32}{x^{3}}+\frac{16}{x^{4}}\)

= \(\frac{16}{x}+\frac{8}{x^{2}}-\frac{32}{x^{3}}+\frac{16}{x^{4}}-4 x+\frac{x^{2}}{2}+\frac{x^{3}}{2}+\frac{x^{4}}{16}-5\).

Question 10.
Find the expansion of (3x² – 2ax + 3a²)³ using binomial theorem.
Answer.
Using Binomial Theorem, the given expression (3x² – 2ax + 3a²)³ can be expanded as [(3x² – 2ax) + 3a²]³

= \({ }^{3} C_{0}\) (3x² – 2ax)3 + \({ }^{3} C_{1}\) (3x² – 2ax)² (3a²) + \({ }^{3} C_{2}\) (3x² – 2ax) (3a²)² + \({ }^{3} C_{3}\) (3a²)³

= (3x² – 2ax)³ + 3 (9x⁴ – 12ax³ + 4a²x²) (3a²) + 3 (3x² – 2ax) (9a⁴) + 27a⁶

= (3x² – 2ax)³ + 81a²x⁴ – 108a³x³ + 36a⁴x² + 81a⁴x² – 54a⁵x + 27a⁶

= (3x² -2ax)³ + 81 a²x⁴ – 108a³x³ + 117a⁴x² – 54a⁵x + 27a⁶ …………….(i)

Again by using Binomial Theorem, we obtain (3x² – 2ax)³
= \({ }^{3} C_{0}\) (3x²)³ – \({ }^{3} C_{1}\) (3x²)² (2ax) + \({ }^{3} C_{2}\) (3x²) (2ax)² – \({ }^{3} C_{3}\)(2ax)³
= 27x⁶ – 3(9x⁴) (2ax) + 3(3x²) (4a²x²) – 8a³x³
= 27x⁶ – 54ax⁵ + 36a²x⁴ – 8a³x³ ………….(ii)

From eqs. (i) and (ii), we obtain (3x² – 2ax + 3a²)³
= 27x⁶ – 54ax⁵ + 36a² x⁴ – 8a³x³ + 81a²x⁴ – 108a³x³ + 117a⁴x² – 54a⁵x + 27a⁶
= 27x⁶ – 54ax⁵ + 117a²x⁴ – 116a³x³ + 117a⁴x² – 54a⁵x + 27a⁶.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 8 Binomial Theorem Ex 8.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.2

Question 1.
Find the coefficient of x⁵ in (x + 3)⁸
Answer.
It is known that (r + 1)th term, (T_{r + 1}), in the binomial expansion of (a + b)ⁿ is given by T_{r + 1} = \({ }^{n} C_{r}\) a^{n – r} b^r
Assuming that x⁵ occurs in the (r + 1)th term of the expansion (x + 3)⁸, we obtain
T_{r + 1} = \({ }^{8} C_{r}\) (x)^{8 – r} (3)^r
Comparing the indices of x in x⁵ and in T_{r + 1}, we obtain r = 3.
Thus, the coefficient of x⁵ is \({ }^{8} C_{3}\) (3)³
= \(\frac{8 !}{3 ! 5 !} \times 3^{3}\)

= \(\frac{8 \cdot 7 \cdot 6 \cdot 5 !}{3 \cdot 2.5 !} \cdot 3^{3}\) = 1512.

Question 2.
Find the coefficient of a⁵ b⁷ in (a – 2b)¹².
Answer.
It is known that (r + 1)th term, (T_{r + 1}), in the binomial expansion of (a + b)ⁿ is given by T_{r + 1} = \(\) a^{n – r} b^r
Assuming that a⁵b⁷ occurs in the (r + 1)th term of the expansion (a – 2b)¹²,we obtain
T_{r + 1} = \({ }^{12} C_{r}\) (a)^{12 – r} (- 2b)^r
= \({ }^{12} C_{r}\) (- 2)^r (a)^{12 – r} (b)^r
Comparing the indices of a and b in a⁵b⁷ and in T_{r + 1}, we obtain r = 7
Thus, the coefficient of a⁵b⁷ is
12C7 (- 2)⁷ = \(\frac{12 !}{7 ! 5 !} \cdot 2^{7}\)
= \(-\frac{12.11 .10 .9 .8 .7 !}{5.4 .3 .2 .7 !} \cdot 2^{7}\)
= – (792) (128)
= – 101376

Question 3.
Write the general term ¡n the expansion of (x² – y)⁶.
Answer.
General term in the expansion of (a + b)ⁿ is given by
T_{r + 1} = \({ }^{n} C_{r}\) a^{n – r} b^r
T_{r + 1} = \({ }^{6} C_{r}\) (x²)^{6 – r} (y)^r
= \({ }^{6} C_{r}\) x^{12 – r} (- y)^r

Question 4.
Write the general term in the expansion of (x² – 2y)¹², x ≠ 0.
Answer.
It is known that the general term T_{r + 1} {which is the (r + 1)th term} in the binomial expansion of (a + b)ⁿ is given by T_{r + 1} = \({ }^{n} C_{r}\) a^{n – r} b^r
Thus, the general term in the expansion of (x² – yx)¹² is
T_{r + 1} = \({ }^{12} C_{r}\) (x²)^{12 – r} (- yx)^r
= (- 1)^r \({ }^{12} C_{r}\) x24-2r . y^r . x^r
= (- 1)^r \({ }^{12} C_{r}\) x^{24 – r} . y^r

Question 5.
Find the 4th term in the expansion of (x – 2y)¹².
Answer.
It is known that (r + 1)th term, (T_{r + 1}), in the binomial expansion of (a + b)ⁿ is given by T_{r + 1} = \({ }^{n} C_{r}\) a^{n – r} b^r
Thus, the 4th term in the expansion of (x – 2y)¹² is
T₄ = T_{3 + 1}
= \({ }^{12} C_{3}[/latex (x)^{12 – 3} (- 2y)³
= (- 1)³ . [latex]\frac{12 !}{3 ! 9 !}\) . x⁹ . (2)³ . y³
= – \(\frac{12 \cdot 11 \cdot 10}{3.2}\) . x⁹ . (2)³ . y³
= – 1760 x⁹ . x⁹ . y³

Question 6.
Find the 13th term in the expansion of (9x – \(\frac{1}{3 \sqrt{x}}\))¹⁸, x ≠ 0.
Answer.

Question 7.
Find the middle terms in the expansions of \(\left(3-\frac{x^{3}}{6}\right)^{7}\).
Answer.
It is known that in the expansion of (a + b)ⁿ, if n is odd, then there are two middle terms, namely \(\left(\frac{n+1}{2}\right)^{t h}\) term and \(\left(\frac{n+1}{2}+1\right)^{t h}\) term.

Therefore, the middle terms in the expansion of \(\left(3-\frac{x^{3}}{6}\right)^{7}\) are \(\left(\frac{7+1}{2}\right)^{t h}\)

Thus, the middle terms in the expansion of \(\left(3-\frac{x^{3}}{6}\right)^{7}\) are – \(\frac{105}{8}\) x⁹ and \(\frac{35}{48}\) x¹².

Question 8.
Find the middle terms in the expansions of \(\left(\frac{x}{3}+9 y\right)^{10}\).
Answer.
It is known that in the expansion (a + b)ⁿ, if n is even, then the middle term is \(\left(\frac{n}{2}+1\right)^{t h}\) term.

Therefore, the middle term in the expansion of \(\left(\frac{x}{3}+9 y\right)^{10}\) is 61236 x⁵ y⁵.

Question 9.
In the expansion of (1 + a)^{m + n}, prove that coefficients of a^m and aⁿ are equal.
Answer.
It is known that (r + 1)th term, (T_{r + 1}), in the binomial expansion of (a + b)ⁿ is given by
T_{r + 1} = \(\) a^{n – r} b^r.

Question 10.
The coefficients of the (r – 1)th, rth and (r + 1)th terms in the expansion of (x + 1)ⁿ are in the ratio 1 : 3 : 5. Find n and r.
Answer.

Question 11.
Prove that the coefficient of xⁿ in the expansion of (1 + x)²ⁿ is twice the coefficient of xⁿ in the expansion of (1 + x)^{2n – 1}.
Answer.

From eqs. (i) and (ii), it is observed that \(\frac{1}{2}{ }^{2 n} C_{n}={ }^{2 n-1} C_{n}\).

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

Therefore, the coefficient of xⁿ in the expansion of (1 + x)²ⁿ is twice the coefficient of xⁿ in the expansion of (1 + x)^{2n – 1}.
Hence proved.

Question 12.
Find a positive value of m for which the coefficient of xⁿ in the expansion (1 + x)^m is 6.
Answer.
It is known that (r + 1)th term, (T_{r + 1}), in the binomial expansion of (a + b)ⁿ is given by T_{r + 1} = \(\) a^{n – r}b^r.

Assuming that x² occurs in the (r + 1)th term of the expansion (1 + x)^m, we obtain
\(T_{r+1}={ }^{m} C_{r}(1)^{m-r}(x)^{r}={ }^{m} C_{r}(x)^{r}\)

Comparing the indices of x in x² and in T_{r + 1}, we obtain r = 2
Therefore, the coefficient of x₂ is \({ }^{m} C_{2}\)

It is given that the coefficient of x² in the expansion (1 + x)^m is 6.

\({ }^{m} C_{2}\) = 6

⇒ \(\frac{m !}{2 !(m-2) !}\) = 6

⇒ \(\frac{m(m-1)(m-2) !}{2 \times(m-2) !}\) = 6
⇒ m (m – 1) = 12
⇒ m² – m – 12 = 0
⇒ m² – 4m + 3m – 12 = 0
⇒ m(m – 4) + 3(m – 4) = 0
⇒ (m – 4) (m + 3) = 0
⇒ (m – 4) = 0 or (m + 3) = 0
⇒ m = 4 or m = – 3.
Thus, the positive value of m, for which the coefficient of x² in the expansion (1 + x)^m is 6, is 4.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 8 Binomial Theorem Ex 8.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 8 Binomial Theorem Ex 8.1

Question 1.
Expand the expression (1 – 2x)⁵.
Answer.
Here, (1 – 2x)⁵ = [1 + (- 2x)]⁵
= \({ }^{5} C_{0}\) + \({ }^{5} C_{1}\) (-2X) + \({ }^{5} C_{2}\) (-2x)2 + \({ }^{5} C_{3}\) (-2x)3 + \({ }^{5} C_{4}\) (-2x)4 + \({ }^{5} C_{5}\) (-2x)s
= 1 + 5 (- 2x) + 10 (4x²) + 10 (- 8x³) + 5(16x⁴) + 1 (- 32x⁵)
= 1 – 10x + 40x² – 80x³ + 80x⁴ – 32x⁵
which is the required expansion.

Question 2.
Expand the expression \(\left(\frac{2}{x}-\frac{x}{2}\right)^{5}\).
Answer.
By using Binomial Theorem, the expression \(\left(\frac{2}{x}-\frac{x}{2}\right)^{5}\) can be expanded as

Question 3.
Expand the expression (2x – 3)⁶.
Answer.
By using Binomial Theorem, the expression (2x – 3)⁶ can be expanded as
(2x – 3) = \({ }^{6} C_{0}\) (2x)⁶ – \({ }^{6} C_{1}\) (2x)⁵ (3) + \({ }^{6} C_{2}\) (2x)⁴ (3)² – \({ }^{6} C_{3}\) (2x)³ (3)³ + \({ }^{6} C_{4}\) (2x)² (3)⁴ – \({ }^{6} C_{5}\) (2x) (3)⁵ + \({ }^{6} C_{6}\) (3)⁶
= 64x⁶ – 6(32x⁵) (3) + (15) (16x⁴) (9) – 20(8x³) (27) + 15(4x²) (81) – 6 (2x) (243) + 729
= 64x⁶ – 576x⁵ + 2160x⁴ – 4320x³ + 4860x² – 2916x + 729

Question 4.
Expand the expression \(\left(\frac{x}{3}+\frac{1}{x}\right)^{5}\).
Answer.
Using Binomial Theorem,

\((a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots+{ }^{n} C_{n} b^{n}\)

Question 5.
Expand: \(\left(x+\frac{1}{x}\right)^{6}\).
Answer.

Question 6.
Using Binomial Theorem, evaluate (96)³.
Answer.
96 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial Theorem can be applied.
It can be written that, 96 = 100 – 4
∴ (96)³ = (100 – 4)³
= \({ }^{3} C_{0}\) (100)³ – \({ }^{3} C_{1}\) (100)² (4) + \({ }^{3} C_{2}\) (100) (4)² – 3C\({ }^{3} C_{0}\) (4)³
= (100)³ – 3(100)² (4) + 3(100) (4)² – (4)³
= 1000000 – 120000 + 4800 – 64 = 884736.

Question 7.
Using Binomial Theorem, evaluate (102)⁵.
Answer.
102 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial Theorem can be applied.
It can be written that, 102 = 100 + 2
∴ (102)⁵ = (100 + 2)⁵

= \({ }^{5} C_{0}\) (100)⁵ + \({ }^{5} C_{1}\) (100)⁴ (2) + \({ }^{5} C_{2}\) (100)³ (2)² + \({ }^{5} C_{3}\) (100)² (2)³ + \({ }^{5} C_{4}\) (100) (2)⁴ + \({ }^{5} C_{5}\) (2)⁵

= (100)⁵ + 5(100)⁴ (2) + 10 (100)³ (2)² + 10 (100)² (2)³ + 5(100) (2)⁴ + (2)⁵
= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32 = 11040808032

Question 8.
Using Binomial Theorem, evaluate (101)⁴.
Answer.
101 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then,
Binomial Theorem can be applied.
If can be written that, 101 = 100 + 1
∴ (101)⁴ =(100 + 1)⁴
= \({ }^{4} C_{0}\) (100)⁴ + \({ }^{4} C_{1}\)(100)³ (1) + \({ }^{4} C_{2}\) (100)² (1)² + \({ }^{4} C_{3}\) (100)(1)³ + \({ }^{4} C_{4}\) (1)⁴.

= (100)⁴ + 4(100)³ + 6(100)² + 4(100) + (1)⁴
= 100000000 + 4000000 + 60000 + 400 +1 =104060401.

Question 9.
Using Binomial Theorem, evaluate (99)⁵.
Answer.
99 can be written as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial Theorem can be applied.
It can be written that, 99 = 100 – 1
(99)⁵ =(100 – 1)⁵
= \({ }^{5} C_{0}\) (1oo)⁵ – \({ }^{5} C_{1}\) (100)⁴ (1) + \({ }^{5} C_{2}\) (100)³ (1)² – \({ }^{5} C_{3}\) (100)² (1)³ + \({ }^{5} C_{4}\) (100) (1)⁴ – \({ }^{5} C_{5}\) (1)⁵

= (100)⁵ – 5(100)⁴ + 10(100)³ – 10 (100)² + 5 (100) – 1
= 10000000000 – 500000000 + 10000000 – 100000 + 500 – 1
= 10010000500 – 500100001 = 9509900499.

Question 10.
Using Binomial Theorem, indicate which number is larger (1.1)10000 or 1000.
Answer.
By splitting 1.1 and then applying Binomial Theorem, the first few terms of (1.1)¹⁰⁰⁰⁰ can be obtained as (1.1)¹⁰⁰⁰⁰ = (1 + 0.1)¹⁰⁰⁰⁰
= \({ }^{10000} C_{0}\)+ \({ }^{10000} C_{1}\) (1.1) + Other positive terms
= 1 + 10000 × 1.1 + Other positive terms
= 1 + 11000 + Other positive terms > 1000
Hence, (1.1)¹⁰⁰⁰⁰ > 1000.

Question 11.
Find (a + b)⁴ – (a – b)⁴.
Hence, evaluate (√3 + √2)⁴ – (√3 – √2)⁴.
Answer.
Using Binomial Theorem, the expression, (a + b)⁴ and (a – b)⁴, can be expanded as
(a + b)⁴ = \({ }^{4} C_{0}\) a⁴ + \({ }^{4} C_{1}\)a³b + \({ }^{4} C_{2}\) a²b² + \({ }^{4} C_{3}\) ab³ + \({ }^{4} C_{4}\) b⁴

(a – b)⁴ = \({ }^{4} C_{0}\) a⁴ – \({ }^{4} C_{1}\)a³b + \({ }^{4} C_{2}\) a²b² – \({ }^{4} C_{3}\) ab³ + \({ }^{4} C_{4}\) b⁴

∴ (a + b)⁴ – (a + b)⁴ =
= \({ }^{4} C_{0}\) a⁴ + \({ }^{4} C_{1}\) a³b + \({ }^{4} C_{2}\) a²b² + \({ }^{4} C_{3}\) ab³ + \({ }^{4} C_{4}\) b⁴ – [ \({ }^{4} C_{0}\) a⁴ – \({ }^{4} C_{1}\) a³b + \({ }^{4} C_{2}\) a²b² – \({ }^{4} C_{3}\) ab³ + \({ }^{4} C_{4}\) b⁴]

= 2 (\({ }^{4} C_{1}\) a³b + \({ }^{4} C_{1}\) ab³)
= 2(4a³b + 4ab³)
= 8ab(a² + b²)
By putting a = √3 and b = √2 , we obtain ,
(√3 + √2)⁴ – (√3 – √2)⁴ = 8(√3) (√2) {(√3)² + ((√2)²)
= 8(√6) {3 + 2} = 40√6.

Question 12.
Find (x + 1)⁶ + (x – 1)⁶. Hence or otherwise evaluate (√2 + 1)⁶ + (√2 – 1)⁶.
Answer.
We have,

(x + 1)⁶ = x⁶ – 6x⁵ + 15x⁴ + 20x³ + 15x² + 6x + 1 ……………..(i)
Similarly, (x – 1)⁶ = x⁶ – 6x⁵ + 15x⁴ – 20x³ + 15x² – 6x + 1 ………….(ii)
Now, adding eqs. (i) and (ii), we get
(x + 1)⁶ + (x – 1)⁶ = 2 [x⁶ + 15x⁴ + 15x² + 1]
Now, putting x = V2, we get
(√2 + 1)⁶ + (√2 – 1)⁶ = 2[(V2)⁶ +15(V2)⁴ + 15(V2)² + 1]
= 2 [2³ + 15 × 2² + 15 × 2 + 1]
= 2(8 + 15 × 4 + 30 + 1]
= 2 [8 + 60 + 30 + 1]
= 2 [99] = 198.

Question 13.
Show that 9^{N + 1} – 8n – 9 is divisible by 64, whenever n is a positive integer.
Answer.
In order to show that 9^{n + 1} – 8n – 9 is divisible by 64, it has to be proved that,
9^{n + 1} – 8n – 9 = 64k, where k is some natural number.
By Binomial Theorem,

⇒ 9^{n + 1} – 8n – 9 = 64k, where k
= \({ }^{n+1} C_{2}+{ }^{n+1} C_{3} \times 8+\ldots+{ }^{n+1} C_{n+1}(8)^{n-1}\) is a natural number.
Thus, 9^{n + 1} – 8n – 9,is divisible by 64, wherever n is a positive integer.

Question 14.
Prove that \(\sum_{r=0}^{n} \mathbf{3}^{r}{ }^{n} C_{r}\) = 4ⁿ.
Answer.
By Binomial Theorem, \(\sum_{r=0}^{n}{ }^{n} C_{r} a^{n-r} b^{r}\) = (a + b)ⁿ
By putting b = 3^and a = 1 in the above equation, we obtain
\(\sum_{r=0}^{n}{ }^{n} C_{r}(1)^{n-r}(3)^{r}\) = (1 + 3)ⁿ
⇒ \(\sum_{r=0}^{n} 3^{r}{ }^{n} C_{r}\) = 4ⁿ
Hence, proved.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 7 Permutations and Combinations Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Miscellaneous Exercise

Question 1.
How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER?
Answer.
There are 8 letters in the word ‘DAUGHTER’ including 3 vowels and 5 consonants.
We have to select 2 vowels out of 3 vowels and 3 consonants out of 5 consonants.
∴ Number of ways of selection = \({ }^{3} C_{2} \times{ }^{5} C_{3}\) = 3 × 10 = 30
Now, each word contains 5 letters which can be arranged among themselves in 5! ways.
So, total number of words = 5 ! x 30 = 120 × 30 = 3600

Question 2.
How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?
Answer.
The word EQUATION consists of 5 vowels and 3 consonants.
∴ 5 vowels can be arranged in 5! = 120 ways.
3 consonants can be arranged in 3! = 6 ways
The two blockof vowels and consonants can be arranged in 2! = 2 ways.
∴ The no. of world which can be formed with 1 letters of the word EQUATION so that vowels and consonants occur together = 120 × 6 × 2 = 1440.

Question 3.
A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:
(i) exactly 3 girls?
(ii) atleast 3 girls?
(iii) atmost 3 girls?
Answer.
A committee of 7 has to be formed from 9 boys and 4 girls.
(i) Since exactly 3 girls are to be there in every committee, each committee must consist of (7 – 3) = 4 boys only.
Thus, in this case, required number of ways = \({ }^{4} C_{3} \times{ }^{9} C_{4}\)
= \(\frac{4 !}{3 ! 1 !} \times \frac{9 !}{4 ! 5 !}\)
= \(4 \times \frac{9 \times 8 \times 7 \times 6 \times 5 !}{4 \times 3 \times 2 \times 1 \times 5 !}\) = 504.

(ii) Since atleast 3 girls are to be there in every committee, the committee can consist of
(a) 3 girls and 4 boys or
(b) 4 girls and 3 boys
3 girls and 4 boys can be selected in \({ }^{4} C_{3} \times{ }^{9} C_{4}\) ways.
4 girls and 3 boys can be selected in \({ }^{4} \mathrm{C}_{4} \times{ }^{9} \mathrm{C}_{3}\) ways.
Therefore, in this case, required number of ways
= \({ }^{4} C_{3} \times{ }^{9} C_{4}\) + \({ }^{4} \mathrm{C}_{4} \times{ }^{9} \mathrm{C}_{3}\)
= 504 + 84 = 588.

(iii) Since atmost 3 girls are to be there in every committee, the committee can consist of
(a) 3 girls and 4 boys
(b) 2 girls and 5 boys
(c) 1 girl and 6 boys
(d) No girl and 7 boys
3 girls and 4 boys can be selected in \({ }^{4} C_{3} \times{ }^{9} C_{4}\) ways.
2 girls and 5 boys can be selected in \({ }^{4} C_{2} \times{ }^{9} C_{5}\) ways.
1 girl and 6 boys can be selected in \({ }^{4} C_{1} \times{ }^{9} C_{6}\) ways.
No girl and 7 boys can be selected in \({ }^{4} C_{0} \times{ }^{9} C_{7}\) ways.
Therefore, in this case, required number, of ways :
= \({ }^{4} \mathrm{C}_{3} \times{ }^{9} \mathrm{C}_{4}+{ }^{4} \mathrm{C}_{2} \times{ }^{9} \mathrm{C}_{5}+{ }^{4} \mathrm{C}_{1} \times{ }^{9} \mathrm{C}_{6}+{ }^{4} \mathrm{C}_{0} \times{ }^{9} \mathrm{C}_{7}\)

= \(\frac{4 !}{3 ! 1 !} \times \frac{9 !}{4 ! 5 !}+\frac{4 !}{2 ! 2 !} \times \frac{9 !}{5 ! 4 !}+\frac{4 !}{1 ! 3 !} \times \frac{9 !}{6 ! 3 !}+\frac{4 !}{0 ! 4 !} \times \frac{9 !}{7 ! 2 !}\)

= 504 + 756 + 336 + 36 = 1632.

Question 4.
If the different permutations of all the letter of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E?
Answer.
Words starting with A are formed with the letters 2I’s, 2N’s, A, E, X, M, T, O.
Number of words formed by these letters = \(\frac{10 !}{2 ! 2 !}\)

= \(\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 2}\) = 907200
Then the words starting with E, I, M, N, O, T, X will be formed.
∴ Number of words before the first word starting with E is formed = 907200.

Question 5.
How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated?
Answer.
A number is divisible by 10 if its units digits is 0.
Therefore, 0 is fixed at the units place.
Therefore, there will be as many ways as there are ways of filling 5 vacant places in succession by the remaining 5 digits (i.e., 1, 3, 5, 7 and 9).
The 5 vacant places can be filled in 5! ways.
Hence, required number of 6-digit numbers = 5! = 120.

Question 6.
The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet?
Answer.
2 different vowels and 2 different consonants are to be selected from the English alphabet.
Since there are 5 vowels in the English alphabet, number of ways of selecting 2 different vowels from the alphabet
= \({ }^{5} C_{2}=\frac{5 !}{2 ! 3 !}\) = 10
Since there are 21 consonants in the English alphabet, number of ways of selecting 2 different consonants from the alphabet
= \({ }^{21} \mathrm{C}_{2}=\frac{21 !}{2 ! 19 !}\)
= 210
Therefore, number of combinations of 2 different vowels and 2 different consonants = 10 × 210 = 2100.
Each of these 2100 combinations has 4 letters, which can be arranged among themselves in 4! ways.
Therefore, required number of words = 2100 × 4! = 50400.

Question 7.
In an examination, a question paper consists of 12 questions divided into two parts i.e., Part I and Part H, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions?
Answer.
Student may select 8 questions according to following scheme

= 10 × 7 + 5 × 35 + 5 × 35
= 70 + 175 +175 = 420 ways.

Question 8.
Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.
Answer.
From a deck of 52 cards, 5-card combinations have to be made in such a way that in each selection of 5 cards, there is exactly one king.
In a deck of 52 cards, there are 4 kings.
1 king can be selected out of 4 kings in \({ }^{4} \mathrm{C}_{1}\) ways.
4 cards out of the remaining 48 cards can be selected in \({ }^{48} \mathrm{C}_{4}\) ways.
Thus, the required number of 5-card combinations is \({ }^{4} \mathrm{C}_{1} \times{ }^{48} \mathrm{C}_{4}\).

Question 9.
It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?
Answer.
5 men and 4 women are to be seated in a row such that the women occupy the even places.
The 5 men can be seated in 5! ways.
For each arrangement, the 4 women can be seated only at the cross marked places (so that women occupy the even places).
M × M × M × M × M
Therefore, the women can be seated in 4! ways.
Thus, possible number of arrangements = 4! × 5!
= 24 × 120 = 2880.

Question 10.
From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen?
Answer.
There are two cases :
(a) If the 3 students join the excursion party then the number of combinations will be P₁ = C (22, 7).
(b) If the 3 students do not join the excursion party. Then the number of combinations P₂ = C (22, 10).
If P is the combination of choosing the excursion party, then
P = P₁ + P₂
= C(22, 7) + C(22,10)
= \(\frac{22 !}{7 ! 15 !}+\frac{22 !}{10 ! 12 !}\)

= \(\frac{22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16 \times 15 !}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 15 !}\) + \(\frac{22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 !}{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 12 !}\)
= 170544 + 646646 = 817190.

Question 11.
In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together?
Answer.
In the given word ASSASSINATION, the letter A appears 3 times, S appears 4 times, I appears 2 times, N appears 2 times, and all the other letters appear only once.
Since all the words have to be arranged in such a way that all the Ss are together, SSSS is treated as a single object for the time being.
This single object together with the remaining 9 objects will account for 10 objects.
These 10 objects in which there are 3 As, 2 Is, and 2 Ns can be arranged in \(\frac{10 !}{3 ! 2 ! 2 !}\) ways.
Thus, required number of ways of arranging the letters of the given word = \(\frac{10 !}{3 ! 2 ! 2 !}\)
= \(\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 !}{3 ! \times 2 \times 2}\)
= 151200.

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 7 Permutations and Combinations Ex 7.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.4

Question 1.
If (\({ }^{n} \mathbf{C}_{8}={ }^{n} C_{2}\)), find (\({ }^{n} \mathbf{C}_{2}\)).
Answer.
It is known that, (\({ }^{n} \mathbf{C}_{a}={ }^{n} C_{2}\))
⇒ a =
⇒ n = a +
Therefore, (\({ }^{n} \mathbf{C}_{8}={ }^{n} C_{2}\))
⇒ n = 8 + 2 = 10
∴ \({ }^{n} \mathbf{C}_{2}={ }^{10} C_{2}\)

= \(\frac{10 !}{2 !(10-2) !}=\frac{10 !}{2 ! 8 !}\)

= \(\frac{10 \times 9 \times 8 !}{2 \times 1 \times 8 !}\) = 45.

Question 2.
Determine n if
(i) (\(\left.{ }^{2 n} C_{3}:{ }^{n} C_{3}\right)\)) = 12 : 1

(ii) \(\left.{ }^{2 n} C_{3}:{ }^{n} C_{3}\right)\)) = 11 : 1
Answer.
(i) Given, (\(\left.{ }^{2 n} C_{3}:{ }^{n} C_{3}\right)\)) = 12 : 1

(ii) \(\left.{ }^{2 n} C_{3}:{ }^{n} C_{3}\right)\)) = 11 : 1

or \(\frac{2 n(2 n-1)(2 n-2)}{1 \times 2 \times 3} \div \frac{n(n-1)(n-2)}{1 \times 2 \times 3}=\frac{11}{1}\)

or \(\frac{4 n(n-1)(2 n-1)}{6} \times \frac{6}{n(n-1)(n-2)}=\frac{11}{1}\)

or 4 (2n – 1) = 11 (n – 2)
or 8n – 4 = 11n – 22
∴ 3n = 18
or n = 6.

Question 3.
How mpny chords can be drawn through 21 points on a circle?
Answer.
For drawing one chord on a circle, only 2 points are required.
To know the number of chords that can be drawn through the given 21 points on a circle, the number of combinations have to be counted.
Therefore, there will be as many chords as there are combinations of 21 points taken 2 at a time.
Thus, required number of chords = \({ }^{21} C_{2}=\frac{21 !}{2 !(21-2) !}\)

= \(\frac{21 !}{2 ! 19 !}=\frac{21 \times 20}{2}\)

= 210.

Question 4.
In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
Answer.
There are 5 boys and 4 girls. We have to select 3 out of 5 boys and 3 out of 4 girls.
∴ Number of ways of selection = \({ }^{5} C_{3} \times{ }^{4} C_{3}\)

= \(\frac{5 !}{3 ! 2 !}=\frac{4 !}{3 ! 1 !}\)

= \(\frac{5 \times 4}{2 \times 1} \times \frac{4}{1}\)
= 10 × 4 = 40.

Question 5.
Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.
Answer.
There are a total of 6 red balls, 5 white balls, and 5 blue balls.
9 balls have to be selected in such a way that each selection consists of 3 balls of each colour.
Here, 3 balls can be selected from 6 red balls in \({ }^{6} \mathrm{C}_{3}\) ways.
3 balls can be selected from 5 white balls in \({ }^{5} \mathrm{C}_{3}\) ways.
3 balls can be selected from 5 blue balls in \({ }^{5} \mathrm{C}_{3}\) ways.
Thus, by multiplication principle, required number of ways of selecting 9 balls ,
= \({ }^{6} \mathrm{C}_{3} \times{ }^{5} \mathrm{C}_{3} \times{ }^{5} \mathrm{C}_{3}\)

= \(\frac{6 !}{3 ! 3 !} \times \frac{5 !}{3 ! 2 !} \times \frac{5 !}{3 ! 2 !}\)

= \(\frac{6 \times 5 \times 4 \times 3 !}{3 ! \times 3 \times 2} \times \frac{5 \times 4 \times 3 !}{3 ! \times 2 \times 1} \times \frac{5 \times 4 \times 3 !}{3 ! \times 2 \times 1}\)
= 20 × 10 × 10 = 2000.

Question 6.
Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.
Answer.
In a deck of 52 cards, there are 4 aces. A combination of 5 cards have to be made in which there is exactly one. ace.
Then, one ace can be selected in \({ }^{4} \mathrm{C}_{1}\) ways and the remaining 4 cards can be selected out of the 48 cards in \({ }^{48} \mathrm{C}_{4}\)ways.
Thus, by multiplication principle, required number of 5 card combinations
= \({ }^{48} C_{4} \times{ }^{4} C_{1}\)

= \(\frac{48 !}{4 ! 44 !} \times \frac{4 !}{1 ! 3 !}\)

= \(\frac{48 \times 47 \times 46 \times 45}{4 \times 3 \times 2 \times 1} \times 4\) = 778320.

Question 7.
In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?
Answer.
Out of 17 players, 5 players are bowlers.
A cricket team of 11 players is to be selected in such a way that there are exactly 4 bowlers.
4 bowlers can be selected in \({ }^{5} \mathrm{C}_{4}\) ways and the remaining 7 players can be selected out of the 12 players in \({ }^{12} \mathrm{C}_{7}\) ways.
Thus, by multiplication principle, required number of ways of selecting cricket team
= \({ }^{5} \mathrm{C}_{4} \times{ }^{12} \mathrm{C}_{7}\)

= \(\frac{5 !}{4 ! 1 !} \times \frac{12 !}{7 ! 5 !}\)

= 5 × \(\frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1}\)

= 3690.

Question 8.
A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.
Answer.
There are 5 black and 6 red balls in the bag.
2 black bails can be selected out of 5 black balls in \({ }^{5} \mathrm{C}_{2}\) ways and 3 red balls can be selected out of 6 red balls in \({ }^{6} \mathrm{C}_{3}\) ways.
Thus, by multiplication principle, required number of ways of selecting 2 black and 3 red balls = \({ }^{5} C_{2} \times{ }^{6} C_{3}\)

= \(\frac{5 !}{2 ! 3 !} \times \frac{6 !}{3 ! 3 !}\)

= \(\frac{5 \times 4}{2} \times \frac{6 \times 5 \times 4}{3 \times 2 \times 1}\)

= 10 × 20 = 200.

Question 9.
In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?
Answer.
There are 9 courses available out of which, 2 specific courses are compulsory for every student.
Therefore, every student has to choose 3 courses out of the remaining 7 courses.
This can be chosen in \(\) ways.
Thus, required number of ways of choosing the programme.

= \({ }^{7} \mathrm{C}_{3}=\frac{7 !}{3 ! 4 !}\)

= \(\frac{7 \times 6 \times 5 \times 4 !}{3 \times 2 \times 1 \times 4 !}\) = 35.