OP Malhotra Class 11 Maths Solutions Chapter 14 Sequence and Series Chapter Test

Effective Class 11 OP Malhotra Solutions Chapter 14 Sequence and Series Chapter Test can help bridge the gap between theory and application.

S Chand Class 11 ICSE Maths Solutions Chapter 14 Sequence and Series Chapter Test

Question 1.
Write down the first five terms of the sequence, whose nth term is (-1)^n-1 . 5ⁿ⁺¹
Solution:
Given T_n = (-1)^n-1 . 5ⁿ⁺¹
∴ T₁ = (-1)^1-1 5¹⁺¹ = 25;
T₂ = (-1) 5³ = – 125 ;
T₃ = 5³⁺¹ = 625 ;
T₄ = -5⁵ = -3125 ;
T₅ = 5⁶ = 15625

Question 2.
If the 3rd and the 6th terms of an A.P. and 7 and 13 respectively, find the first term and the common difference.
Solution:
Let a be the first term and d be the common difference of an A.P.
Given T₃ = 7 ⇒ a + 2d = 7 …(1)
[∵ T_n = a + (n – 1)d]
and T₆ = 13 ⇒ a + 5d = 13 …(2)
eqn. (2) – eqn. (1) gives; 3d = 6 ⇒ d = 2
∴ from (1); a + 4 = 7 ⇒ a = 3
Hence the required first term of given A.P be 3 and common difference be 2 .

Question 3.
Find the sum of all natural numbers between 100 and 1000 which are multiple of 5.
Solution:
The natural numbers between 100 and 1000 which are multiple of 5 are 105, 110, 115, ….., 995
Clearly sequence forms A.P with first term a = 105 and common difference = d = 110 – 105 = 5
Let n be the required no. of terms s.t T_n = 995
995 = a+ (n – 1) d ⇒ 995 = 105 + (n – 1) 5
⇒ 890 = (n – 1) 5
⇒ n – 1 = \(\frac { 890 }{ 5 }\) = 178
⇒ n = 179
∴ required sum = S₁₇₉ = \(\frac { 179 }{ 2 }\) [105 + 995]
[∵ S_n = \(\frac { n }{ 2 }\) (a + l)]
= \(\frac { 179 }{ 2 }\) × 1100 = 179 × 550 = 98450

Question 4.
How many terms of the A.P., -6, \(\frac { -11 }{ 2 }\), -5, ……. are needed to give the sum -25 ?
Solution:
Here first term a = – 6 and d = common difference =-\(\frac { 11 }{ 2 }\) + 6 = \(\frac { 1 }{ 2 }\)
Let n be the required number of terms s.t S_n = -25

Hence the required no. of terms of A.P are 5 or 20.

Question 5.
Determine the sum of the first 35 terms of an A.P. if a₂ = 2 and a₇ = 22.
Solution:
Let a be the first term and d be the common difference of an A.P.
Given a₂ = 2 ⇒ 2 = a + d …(1)
and a₇ = 22 ⇒ 22 = a + 6d …(2)
eqn. (2) – eqn. (1) gives ;
20 = 5d ⇒ d = 4
∴ from (1); a = 2 – 4 = -2
We know that S_n = \(\frac { n }{ 2 }\)[2a + (n – 1)d]
∴ S₃₅ = \(\frac { 35 }{ 2 }\)[- 4 + (35 – 1)4] = \(\frac { 35 }{ 2 }\)[- 4 + 136] = 35 × 66 = 2310

Question 6.
If the first term of an A.P. is 2 and the sum of first five terms is equal to one-fourth of the sum of the next five terms, show that the 200th term is -112.
Solution:
Let a be the first term and d be the common differenceof an A.P.
given a = 2 and S₅ = \(\frac { 1 }{ 4 }\) [S₁₀ – S₅] ⇒ 4S₅ = S₁₀ – S₅ ⇒ 5S₅ = S₁₀
⇒ 5 × \(\frac { 5 }{ 2 }\) [2 × 2 + (5 – 1)d] = \(\frac { 10 }{ 2 }\) [2 × 2 + (10 – 1)d] ⇒ \(\frac { 25 }{ 2 }\)[4 + 4d] = 5 [4 + 9d]
⇒ 50(1 + d) = 5(4 + 9d) ⇒ 10 + 10d = 4 + 9d ⇒ d = – 6
∴ T₂₀ = a + 19d = 2 + 19 × (-6) = 2 – 114 = -112

Question 7.
Insert 3 arithmetic means between 2 and 10.
Solution:
Let A₁, A₂, A₃ are three A.M’s between 2 and 10 .
Then 2, A₁, A₂, A₃, 10 are in A.P.
∴ 10 = T₅ = a + 4d = 2 + 4d ⇒ 8 = 4d ⇒ d = 2
Thus, A₁ = T₂ = a + d = 2 + 2 = 3
A₂ = T₃ = a + 2d = 2 + 4 = 6
A₃ = T₄ = a + 3d = 2 + 6 = 8
Hence the required 3 A.M’s between 2 and 10 are 4, 6 and 8 .

Question 8.
Find 12 th term of a G.P. whose 8th term is 192 and the common ratio is 2 .
Solution:
Let a be the first term and r be the common ratio of G.P. Then r = 2
given T₈ = 192 ⇒ ar⁷ = 192 [∵ T_n = ar^n-1]
⇒ a × 2⁷ = 19 ⇒ a = \(\frac { 192 }{ 128 }\) = \(\frac { 3 }{ 2 }\)
∴ T₁₂ = ar¹¹ = \(\frac { 3 }{ 2 }\) × 2¹¹ = 3 × 2¹⁰ = 3 × 1024 = 3072

Question 9.
The first term of a G.P. is 1. The sum of the third and fifth terms is 90. Find the common ratio of the G.P.
Solution:
Let a be the first term and r be the common ratio of G.P. Then a = 1
given T₃ + T₅ = 90 ⇒ ar² + ar⁴ = 90
⇒ r² + r⁴ = 90 ⇒ r⁴ + r² – 90 = 0
⇒ (r – 3) (r³ + 3r² + 10r + 30) = 0
⇒ (r – 3) (r + 3) (r² + 10) = 0
⇒ r = 3, – 3 [while r² + 10 = 0 does not any real values of r]
Hence the required common ratio of given G.P be ± 3.

Question 10.
The sum of first three terms of a G.P. is \(\frac { 39 }{ 10 }\) and their product is 1 . Find the common ratio and the terms.
Solution:
Let the first three terms of G.P are \(\frac { a }{ r }\), a, ar
where a = first term and r = common ratio given product of these numbers be 1.
∴ \(\frac { a }{ r }\) × a × ar = 1 ⇒ a³ = 1 ⇒ a = 1
and Sum of first three terms be \(\frac { 39 }{ 10 }\)

Question 11.
The sum of some terms of a G.P. is 315 and the first term and the common ratio are 5 and 2 respectively. Find the last term and the number of terms.
Solution:
Let a be the first term and r be the common ratio of G.P respectively
Then a = 5 ; r = 2
Let n be the required number of terms s.t S_n = 315
⇒ \(\frac{a\left(r^n-1\right)}{r-1}\) = 315
⇒ \(\frac{5\left(2^n-1\right)}{2-1}\) = 315
⇒ 2ⁿ – 1 = 63
⇒ 2ⁿ = 64 = 2⁶
⇒ n = 6
Hence the required no. of terms of G.P be 6
and required last term = T_n = ar^n-1 i.e.
T₆ = 5 × 2^6-1 = 5 × 32 = 160

Question 12.
Find the sum of the series 0.6 + 0.66 + 0.666 + …… to the n terms.
Solution:
Let S_n = 0.6 + 0.66 + 0.666 + ….. n terms
= 6[0.1 + 0.11 + 0.111 + ….. n terms]
= \(\frac { 6 }{ 9 }\) [0.9 + 0.99 + 0.999 + …… n terms]
= \(\frac { 6 }{ 9 }\) [(1 – 0.1) + (1 – 0.01) + (1 – 0.001) + …… n terms]

Question 13.
The sum of an infinite series is 15 and the sum of the squares of these terms is 45 . Find the series.
Solution:
Let a be the first term and r be the common ratio of an infinite G.P. given
given S_∞ = 15 ⇒ 15 = \(\frac { a }{ 1 – r }\) …(1)
The sum of squares of these terms is 45 .
∴ a² + a²r² + a²r⁴ + …… ∞ = 45 ⇒ \(\frac{a^2}{1-r^2}\) = 45 …(2)
On squaring eqn. (1); we have
\(\frac{a^2}{(1-r)^2}\) = 225 …(3)
On dividing eqn. (3) by eqn. (2); we get
\(\frac{1}{(1-r)^2}\) × 1 – r² = \(\frac { 225 }{ 45 }\) = 5 ⇒ \(\frac{1+r}{1-r}\) = 5 [∵ r ≠ 1]
⇒ 1 + r = 5 – 5r
⇒ 6r = 4
⇒ r = \(\frac { 2 }{ 3 }\)
∴ from (1); 15 = \(\frac{a}{i-\frac{2}{3}}\) ⇒ 15 = 3a ⇒ a = 5
Thus the required G.P be a, ar, ar², ….
5, 5 × \(\frac { 2 }{ 3 }\), 5 × \(\left(\frac{2}{3}\right)^2\), ….. i.e. 5, \(\frac { 10 }{ 3 }\), \(\frac { 20 }{ 9 }\), …..

Question 14.
Insert three geometric means between 1 and 256.
Solution:
Let G₁, G₂, G₃ are the three G.M’s between 1 and 256 .
Then 1, G₁, G₂, G₃, 256 are in G.P
∴ 256 = T₅ = ar⁴ = r⁴ ⇒ r⁴ = 4⁴ ⇒ r = 4
Thus G₁ = T₂ = ar = 1 × 4 = 4; G₂ = T₃ = ar² = 1 × 4² = 16
and G₃ = T₄ =ar³ = 1 × 4³ = 64
Hence the required three G.M’s between 1 and 256 are 4, 16 and 64.

Question 15.
In the sum to infinity of the series 3 + (3 + x)\(\frac { 1 }{ 4 }\) + (3 + 2x) \(\frac{1}{4^2}\) + ….. is \(\frac { 44 }{ 9 }\), find x.
Solution:
S = 3 + (3 + x)\(\frac { 1 }{ 4 }\) + (3 + 2x) \(\frac{1}{4^2}\) + ….. ∞ ….. (1)
∴ \(\frac { 1 }{ 4 }\)S = \(\frac { 3 }{ 4 }\) + (3 + x) \(\frac{1}{4^2}\) + ….. ∞ ….. (2)
eqn. (1) – eqn. (2) gives;
\(\frac { 3 }{ 4 }\)S = 3 + \(\frac { x }{ 4 }\) + \(\frac{x}{4^2}\) + ….. ∞
= 3 + \(\frac{\frac{x}{4}}{1-\frac{1}{4}}\) = 3 + \(\frac { x }{ 4 }\) × \(\frac { 4 }{ 3 }\) = 3 + \(\frac { x }{ 3 }\)
⇒ S = \(\frac { 4 }{ 3 }\)\(\left[3+\frac{x}{3}\right]\)
Also given S = \(\frac { 44 }{ 9 }\)
∴ \(\frac { 44 }{ 9 }\) = ∴ \(\frac { 4 }{ 3 }\)\(\left[3+\frac{x}{3}\right]\) ⇒ 3 + \(\frac { x }{ 3 }\) = \(\frac { 44 }{ 9 }\) × \(\frac { 3 }{ 4 }\) = \(\frac { 11 }{ 3 }\) ⇒ \(\frac { x }{ 3 }\) = \(\frac { 11 }{ 3 }\) – 3 = \(\frac { 2 }{ 3 }\) ⇒ x = 2

Question 16.
Find teh sum to n terms of the series 3.8 + 6.11 + 9.14 + ….
Solution:
T_n = (nth term of 3, 6, 9, ….) × (nth term of 8, 11, 14, …)
= [3 + (n – 1) 3] [8 + (n – 1)3] = 3n (3n + 5)
∴ T_n = 9n² + 15n
Thus S_n = ΣT_n = 9Σn² + 15Σn = \(\frac{9 n(n+1)(2 n+1)}{6}\) + \(\frac{15 n(n+1)}{2}\)
= \(\frac{3 n(n+1)(2 n+1)}{2}\) + \(\frac{15 n(n+1)}{2}\)
= \(\frac{n(n+1)}{2}\) [3(2n + 1) + 15]
= \(\frac{n(n+1)}{2}\) (6n + 18) = 3n (n + 1) (n + 3)

Question 17.
Find the sum 5² + 6² + 7² + ….. + 20².
Solution:

Question 18.
If in a geometric progression consisting of positive terms, each term equals the sum of the next two terms, then the common ratio of this progression equals
(a) \(\sqrt{5}\)
(b) \(\frac{1}{2}(\sqrt{5}-1)\)
(c) \(\frac{1}{2}(1-\sqrt{5})\)
(d) \(\frac{1}{2} \sqrt{5}\)
Solution:
Let a be the fast term and r be the common ratio of G.P respectively.
it is given that T_n = T_n+1 + T_n+2
⇒ ar^n-1 = arⁿ + arⁿ⁺¹
⇒ r^n-1 = rⁿ + rⁿ⁺¹
∴ r = \(\frac{-1 \pm \sqrt{1-4(-1)}}{2}\) = \(\frac{-1 \pm \sqrt{5}}{2}\)
since given G.P consisting of positive terms
∴ r > 0 ⇒ r = \(\frac{-1+\sqrt{5}}{2}\)
∴ Ans. (b)

Question 19.
If the first term of an infinite G.P. is 1 and each term is twice the sum of the succeeding terms, then the sum of the series is
(a) 2
(b) \(\frac{5}{2}\)
(c) \(\frac{7}{2}\)
(d) \(\frac{3}{2}\)
(e) \(\frac{9}{2}\)
Solution:
Let a be the first term and r be the common ratio of given G.P.
Then a = 1
Also it is given that T_n = 2(T_n+1 + T_n+2 + …..)
⇒ ar^n-1 = 2 [arⁿ + arⁿ⁺¹ + …… ∞]
⇒ r^n-1 = 2 (rⁿ + rⁿ⁺¹ + ….. ∞)
⇒ r^n-1 = \(\frac{2 r^n}{1-r}\)
⇒ 1 = \(\frac{2 r}{1-r}\)
⇒ 1 – r = 2r
⇒ 3r = 1
⇒ r = \(\frac{1}{3}\)
Hence the required G.P becomes a, ar, ar², …… i.e. 1, \(\frac{1}{3}\), \(\frac{1}{9}\), ….. ∞
i.e. S_∞ = 1 + \(\frac{1}{3}\) + \(\frac{1}{9}\) + ….. ∞
= \(\frac{1}{1-\frac{1}{3}}\) = \(\frac{3}{2}\)
∴ Ans. (d)

Question 20.
If fifth term of a G.P. is 2, then the product of its first 9 terms is
(a) 256
(b) 512
(c) 1024
(d) none of these
Solution:
Let a be the first term and r be the common ratio of G.P.
given T₅ = 2 ⇒ ar⁴ = 2 …(1) [∵ T_n = ar^n-1]
∴ product of first 9 terms = a(ar) (ar²) (ar³) (ar⁴) (ar⁵) (ar⁶) (ar⁷) (a r⁸)
\(=a^9 r^{1+2+3+\ldots .+8}=a^9 r^{\frac{8(8+1)}{2}}=a^9 a^{36}=\left(a r^4\right)^9=2^9=512\)
∴ Ans. (b)

Question 21.
The sum of three decreasing numbers in A.P. is 27 . If -1, -1, 3 are added to them respectively, the resulting series is in G.P. The numbers are
(a) 5, 9, 13
(b) 15, 9, 3
(c) 13, 9, 5
(d) 17, 9, 1
Solution:
Let the three numbers in decreasing A.P are a + d, a, a – d and their sum is 27 .
∴ a + d + a + a – d = 27
⇒ 3a = 27
⇒ a = 9
if -1, -1 and 3 are added to these numbers then resulting numbers a + d – 1, a – 1, a – d + 3 forms G.P.
∴ (a – 1)² = (a + d – 1)(a – d + 3)
⇒ (9 – 1)² = (9 + d – 1)(9 – d + 3)
⇒ 64 = (8 + d)(12 – d)
⇒ 64 = -d² + 4d + 96
⇒ d² – 4d – 32 = 0
⇒ d² – 8d + 4d – 32 = 0
⇒ d(d – 8) + 4(d – 8) = 0
⇒ (d + 4)(d – 8) = 0
⇒ d = -4 or 8
When a = 9 and d = -4 then required three numbers are; 9 – 4, 9, 9 + 4 i.e. 5, 9, 13
When a = 9 and d = 8 then required three numbers are ; 9 + 8, 9, 9 – 8 i.e. 17, 9, 1
Hence the required decreasing A.P be 17, 9, 1
∴ Ans. (d)

Question 22.
The first two terms of geometric progression add up to 12 . The sum of the third and the fourth terms is 48 . If the terms of the geometric progression are alternately positive and negative, then the first term is
(a) -4
(b) -12
(c) 12
(d) 4
Solution:
Let a be the first term and r be the common ratio of G.P respectively.
According to given condition, we have
a + ar = 12 ⇒ a(1 + r) = 12 …(1)
and ar² + ar³ = 48 ⇒ ar²(1 + r) = 48 …(2)
On dividing eqn. (2) by eqn. (1) ;
\(\frac{a r^2(1+r)}{a(1+r)}\) = \(\frac{48}{12}\) = 4
⇒ r² = 4
⇒ r = ± 2
Since the terms of G.P are alternatively positive and negative ∴r = – 2
∴ from (1); a(1 – 2) = 12 ⇒ a = – 12
Hence the required first term of G.P be -12∴ Ans. (b)

Question 23.
The sum to infinity of the series 1 + \(\frac { 2 }{ 3 }\) + \(\frac{6}{3^2}\) + \(\frac{10}{3^3}\) + \(\frac{14}{3^4}\) + …. is
(a) 6
(b) 2
(c) 3
(d) 4
Solution:
Let S = 1 + \(\frac { 2 }{ 3 }\) + \(\frac{6}{3^2}\) + \(\frac{10}{3^3}\) + \(\frac{14}{3^4}\) + …. ∞
after leaving first term, given series be an arithmetico geometric series.
∴ \(\frac { 1 }{ 3 }\)S = \(\frac { 1 }{ 3 }\) + \(\frac{2}{3^2}\) + \(\frac{6}{3^3}\) + \(\frac{10}{3^4}\) …. ∞ ….(2)
eqn. (1) – eqn. (2) gives ;
\(\frac { 2 }{ 3 }\)S = 1 + \(\frac { 1 }{ 3 }\) + \(\frac{4}{3^2}\) + \(\frac{4}{3^3}\) + \(\frac{4}{3^4}\) + …. ∞
⇒ \(\frac { 2 }{ 3 }\)S = \(\frac { 4 }{ 3 }\) + \(\left[\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^4}+\ldots \ldots \infty\right]\)
\(\frac { 2 }{ 3 }\)S = \(\frac { 4 }{ 3 }\) + \(\frac{\frac{4}{3^2}}{1-\frac{1}{3}}\) = \(\frac { 4 }{ 3 }\) + \(\frac { 4 }{ 9 }\) × \(\frac { 3 }{ 2 }\) = \(\frac { 4 }{ 3 }\) + \(\frac { 2 }{ 3 }\) = 2
⇒ S = 2 × \(\frac { 3 }{ 2 }\) = 3
∴ Ans. (c)

Question 24.
The sum of all odd numbers between 1 and 1000 which are divisible by 3 , is
(a) 83667
(b) 90000
(c) 83660
(d) None of these
Solution:
The odd numbers between 1 and 1000 which are divisible by 3 are ; 3, 9, 15, 21, …. 99, …. 999
it clearly forms A.P with first term a = 3 and common difference = 9 – 3 = 6 = d
Also l = 999 ⇒ 999 = a + (n – 1) d = 3 + (n – 1) 6
⇒ \(\frac { 996 }{ 6 }\) = (n – 1) ⇒ n – 1 = 166
⇒ n = 167
We know that S_n = \(\frac { n }{ 2 }\)[a +l]
∴ S₁₆₇ = \(\frac { 167 }{ 2 }\) [3 + 999] = \(\frac { 167 }{ 2 }\) × 1002 = 167 × 601 = 83667
∴ Ans. (c)

Question 25.
If a, b, c are in G.P. and x, y are arithmetic means of a, b and b, c respectively, then \(\frac { 1 }{ x }\) + \(\frac { 1 }{ y }\) is equal to
(a) \(\frac { 2 }{ b }\)
(b) \(\frac { 3 }{ b }\)
(c) \(\frac { b }{ 3 }\)
(d) \(\frac { b }{ 2 }\)
(e) \(\frac { 1 }{ b }\)
Solution:
Given a, b, c are in G.P ⇒ b² = ac …(1)
x be the A.M of a, b
∴ x = \(\frac { a+b }{ 2 }\) …(2)
and y be the A.M of b and c
∴ y = \(\frac { b+c }{ 2 }\) …(3)
∴ \(\frac { 1 }{ x }\) + \(\frac { 1 }{ y }\) = \(\frac { 2 }{ a+b }\) + \(\frac { 2 }{ b + c }\) = \(\frac{2(a+c+2 b)}{a b+a c+b c+b^2}\) = \(\frac{2(a+c+2 b)}{a b+a c+b c+a c}\) = \(\frac{2(a+c+2 b)}{a b+b c+2 b^2}\) = \(\frac{2(a+c+2 b)}{b(a+c+2 b)}\) = \(\frac{2}{b}\)
∴ Ans. (a)