OP Malhotra Class 11 Maths Solutions Chapter 14 Sequence and Series Ex 14(f)

Parents can use Class 11 OP Malhotra Solutions Chapter 14 Sequence and Series Ex 14(f) to provide additional support to their children.

S Chand Class 11 ICSE Maths Solutions Chapter 14 Sequence and Series Ex 14(f)

Question 1.
Find the sum to
(i) 8 terms of 3 + 6 + 12 + …..
(ii) 20 terms of 2 + 6 + 18 + ……..
(iii) 10 terms of 1 + √3 + 3 + …….
(iv) n terms of 3\(\frac { 3 }{ 8 }\) + 2\(\frac { 1 }{ 4 }\) + 1\(\frac { 1 }{ 2 }\) + …..
Solution:
(i) Given series 3 + 6 + 12 + ……… 8 terms G.P with first term a = 3
and common ratio r = 2 > 1 and n = 8
∴ S_n = \(\frac{a\left(r^n-1\right)}{r-1}\) = \(\frac{3\left(2^8-1\right)}{2-1}\)
= 3 × 255 = 765

(ii) Given series be, 2 + 6 + 18 + …….
it clearly forms G.P with first term a = 2
and common ratio r = \(\frac { 6 }{ 2 }\) = 3 > 1 ; n = 20
We know that
S_n = \(\frac{a\left(r^n-1\right)}{r-1}\) = \(\frac{2\left(3^{20}-1\right)}{3-1}\) = 3²⁰ – 1

(iii) Given series be 1 + √3 + 3 +……
it clearly form G.P with first term a = 1
and common ratio = r = √3 > 1 and n = 10
We know that

(iv) Given series be, 3\(\frac { 3 }{ 8 }\) + 2\(\frac { 1 }{ 4 }\) + 1\(\frac { 1 }{ 2 }\) + ……
Clearly it forms G.P. with first term a = 3\(\frac { 3 }{ 8 }\) = \(\frac { 27 }{ 8 }\)
and common ratio = r = \(\frac{\frac{9}{4}}{\frac{27}{8}}\) = \(\frac{9}{4}\) × \(\frac{8}{27}\)
∴ r = \(\frac { 2 }{ 3 }\) < 1
We know that S_n = \(\frac{a\left(1-r^n\right)}{1-r}\)
= \(\frac{\frac{27}{8}\left[1-\left(\frac{2}{3}\right)^n\right]}{1-\frac{2}{3}}\) = \(\frac{81}{8}\)\(\left[1-\left(\frac{2}{3}\right)^n\right]\)

Question 2.
Sum the following series to infinitely:
(i) 1 + \(\frac { 1 }{ 2 }\) + \(\frac { 1 }{ 4 }\) + \(\frac { 1 }{ 8 }\) + ….
(ii) 16, -8, 4 ……
(iii) √2 – \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{2 \sqrt{2}}\) – \(\frac{1}{4 \sqrt{2}}\) + ……
(iv) √3 + \(\frac{1}{\sqrt{3}}\) + \(\frac{1}{3 \sqrt{3}}\) …..
Solution:
(i) Given series be, 1 + \(\frac { 1 }{ 2 }\) + \(\frac { 1 }{ 4 }\) + \(\frac { 1 }{ 8 }\) + …. ∞
it clearly forms G.P with a = 1; r = \(\frac { 1 }{ 2 }\) < 1
∴ S_∞ = \(\frac{a}{1-r}\) = \(\frac{1}{1-\frac{1}{2}}\) = 2

(ii) Given series be, 16 + (-8) + 4 + …. it clearly forms G.P with a = 16 and
r = \(\frac{-8}{16}\) = \(\frac{-1}{2}\) < 1
∴ S_∞ = \(\frac{a}{1-r}\) = \(\frac{16}{1-\left(-\frac{1}{2}\right)}\) = \(\frac{16}{\frac{3}{2}}\) = \(\frac{32}{3}\)

(iii) Given series be
\(\sqrt{2}\) – \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{2 \sqrt{2}}\) – \(\frac{1}{4 \sqrt{2}}\) …. ∞
it clearly forms G.P with first term a = √2
and common ratio r = \(\frac{-\frac{1}{\sqrt{2}}}{\sqrt{2}}\) = –\(\frac { 1 }{ 2 }\)
∴ |r| = \(\frac { 1 }{ 2 }\) < 1
Thus S_∞ =\(\frac{a}{1-r}\) = \(\frac{\sqrt{2}}{1+\frac{1}{2}}\) = \(\frac{2 \sqrt{2}}{3}\)

(iv) Given series clearly forms G.P. with first term a = √3
and common ratio = r = \(\frac{1}{\frac{\sqrt{3}}{\sqrt{3}}}\) = \(\frac { 1 }{ 3 }\) < 1
∴ S_∞ = \(\frac{a}{1-r}\) = \(\frac{\sqrt{3}}{1-\frac{1}{3}}\) = \(\frac{3 \sqrt{3}}{2}\)

Question 3.
Find the sum of a geometric series in which a = 16, r = \(\frac { 1 }{ 4 }\), l = \(\frac { 1 }{ 64 }\)
Solution:

Question 4.
Find the sum of the series 81 – 27 + 9 – … – \(\frac { 1 }{ 27 }\)
Solution:
Given series be
81 – 27 + 9 – …. –\(\frac { 1 }{ 27 }\)
it clearly forms G.P with first term a = 81
and common ratio r =\(\frac {-27 }{ 81 }\) = \(\frac { -1 }{ 3 }\)
Also l = – \(\frac { 1 }{ 27 }\)

Question 5.
The first three terms of a G.P. are x, x + 3, x + 9. Find the value of x and the sum of first eight terms.
Solution:
Given x, x + 3 and x + 9 are in G.P.
∴ (x + 3)² = x(x + 9)
⇒ x² + 6x + 9 = x² + 9x
⇒ 3x = 9 ⇒ x = 3
∴ The three terms of G.P are 3, 6, 12
Here a = 3 ; r = \(\frac { 6 }{ 3 }\) = 2 > 1
Thus S₈ = \(\frac{a\left(r^n-1\right)}{r-1}\) = \(\frac{3\left(2^8-1\right)}{2-1}\) = 3 × 255

Question 6.
Of how many terms is \(\frac { 55 }{ 72 }\), the sum of the series \(\frac { 2 }{ 9 }\) – \(\frac { 1 }{ 3 }\) + \(\frac { 1 }{ 2 }\) …. ?
Solution:
Given series forms G.P with first term a = \frac{2}{9}$
and common ratio = r = \(\frac{-\frac{1}{3}}{\frac{2}{9}}\) = \(\frac{-3}{2}\)
Let n be the required no. of terms

Question 7.
The second term of a G.P. is 2 and the sum of infinite terms is 8 . Find the first term.
Solution:
Let a be the first term and r be the common ratio of given G.P.
Given T₂ = 2 ⇒ ar =2 …(1)
and S_∞ = 8 ⇒ \(\frac{a}{1-r}\) = 8
⇒ \(\frac{a}{1-\frac{2}{a}}\) = 8 [using (1)]
⇒ \(\frac{a^2}{a-2}\) = 8 ⇒ a² – 8a + 16 = 0
⇒ (a – 4)² = 0
⇒ a = 4
Hence the required first term of G.P be 4.

Question 8.
(i) Find the value of 0.234 regarding it as a geometric series.
(ii) Evaluate : (a)\(0.9 \overline{7}\) (b) \(0. \overline{45}\) (c) \(0.23 \overline{45}\)
(iii) Find a rational number which when expressed as a decimal will have \(1.2 \overline{56}\) as its expansion.
Solution:

Question 9.
If a + b + …. + l is a G.P., prove that its sum is \(\frac{b l-a^2}{b-a}\).
Solution:
Given a + b + …….. + l forms G.P.
with first term = a; r = \(\frac{b}{a}\) and last term = l
We know that
S_n = \(\frac{a\left(r^n-1\right)}{r-1}\) = \(\frac{a r^{n-1} \cdot r-a}{r-1}\)
⇒ S_n = \(\frac{l \cdot \frac{b}{a}-a}{\frac{b}{a}-1}\) = \(\frac{l b-a^2}{b-a}\)

Question 10.
The nth term of a geometrical progression is \(\frac{2^{2 n-1}}{3}\) for all values of the first three terms and calculate the sum of the first 10 terms, correct to 3 significant figures.
Solution:
Given T_n = \(\frac{2^{2 n-1}}{3}\)
∴ T₁ = a = \(\frac{2^2-1}{3}\) = \(\frac{2}{3}\);
T₂ = \(\frac{2^{4-1}}{3}\) = \(\frac{8}{3}\);
T₃ = \(\frac{2^{5}}{3}\) = \(\frac{32}{3}\);
Clearly a = \(\frac{2}{3}\)
and common ratio = r = \(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\) = \(\frac{\frac{8}{3}}{\frac{2}{8}}\) = 4 > 1
We know that S_n = \(\frac{a\left(r^n-1\right)}{r-1}\)
∴ S₁₀ = \(\frac{\frac{2}{3}\left[4^{10}-1\right]}{4-1}\) = \(\frac{2}{9}\) (4¹⁰ – 1)
= 233016.6667 = 233000 (correct to 3 significant figures)

Question 11.
A geometrical progression of positive terms and an arithmetical progression have the same first term. The sum of their first terms is 1 , the sum of their second terms is \(\frac{1}{2}\) and the sum of their third terms is 2. Calculate the sum of their fourth terms.
Solution:
Let a be the first term of both series A.P and G.P
Let d be the common difference of an A.P and r be the common ratio of given G.P.
Given sum of their first terms = 1
⇒ a + a = 1 ⇒ a = \(\frac{1}{2}\)
Also, second term of A.P + second term of G.P = \(\frac{1}{2}\)
⇒ (a + d) + ar = \(\frac{1}{2}\) ⇒ \(\frac{1}{2}\) + d + \(\frac{r}{2}\) = \(\frac{1}{2}\)
⇒ 2d + r = 0 ….(1)
Also, third term of A.P + third term of G.P = 2
⇒ (a + 2d) + ar² = 2 ⇒ \(\frac{1}{2}\) + 2d + \(\frac{r^2}{2}\) = 2
⇒ 4d + r² = 3 ….(2)
∴ from eqn. (1) and eqn. (2) ; we have
4d + (-2d)² = 3
⇒ 4d² + 4d – 3 = 0
⇒ d = \(\frac{-4 \pm \sqrt{16+48}}{8}\) = \(\frac{-4+8}{8}\) = \(\frac{1}{2}\), –\(\frac{3}{2}\)
When d = \(\frac{1}{2}\) ∴ from (1); r = – 1
which is not possible since G.P is given to be progression of positive terms
∴ d = –\(\frac{3}{2}\) and from (1); r = + 3
∴ required sum of fourth terms of A.P and
G.P = a + 3d + ar³ = \(\frac{1}{2}\) – \(\frac{9}{2}\) + \(\frac{1}{2}\) × 27
= – 4 + \(\frac{27}{2}\) = \(\frac{19}{2}\)

Question 12.
In a geometric progression, the third term exceeds the second by 6 and the second exceeds the first by 9 . Find
(i) the first term, (ii) the common ratio and (iii) the sum of the first ten terms.
Solution:
Let a be the first term and r be the common ratio of G.P.
Given T₃ = T₂ + 6 ⇒ ar² = ar + 6
⇒ ar₂ – ar = 6 ….(1)
and T₂ = T₁ + 9 ⇒ ar = a + 9
⇒ ar – a = 9 …(2)
On dividing eqn. (1) by eqn. (2); we have
\(\frac{r^2-r}{r-1}\) = \(\frac{6}{9}\) = \(\frac{2}{3}\) ⇒ 3r² – 3r – 2r – 2
⇒ 3r² – 5r + 2 = 0
⇒ (r – 1) (3r – 2) = 0
⇒ r = 1, \(\frac{2}{3}\)
when r = 1 ∴ from (1); a – a = 6 ⇒ 0 = 6
which is false.
Thus r = \(\frac{2}{3}\) ∴ from (1); we have
\(a\left(\frac{2}{3}-1\right)=9\)
⇒ \(a\left(-\frac{1}{3}\right)=9\)
⇒ a = – 27
We know that S_n = \(\frac{a\left(1-r^n\right)}{1-r}\)
∴ S₁₀ = \(\frac{-27\left[1-\left(\frac{2}{3}\right)^{10}\right]}{1-\frac{2}{3}}\) = – 81 \(\left[1-\left(\frac{2}{3}\right)^{10}\right]\)

Question 13.
In an infinite geometric progression, the sum of first two terms is 6 and every term is four times the sum of all the terms that follow it. Find :
(i) the geometric progression and
(ii) its sum to infinity.
Solution:
Let a be the first term and r be the common ratio of G.P.
given a + ar = 6 ….(1)
given T_n = = 4 [T_n+1 T_n+2+…..]
or ar^n-1 = 4[arⁿ + arⁿ⁺¹ … ∞]
⇒ ar^n-1 = 4arⁿ[1 + r + r² … ∞]
⇒ ar^n-1 = \(\frac{4 a r^n}{1-r}\) ⇒ (1 – r)r^n-1 = 4rⁿ
⇒ (1 – r) = 4r ⇒ 5r = 1 ⇒ r = \(\frac{1}{5}\)
∴ from (1) ; a\(\left(1+\frac{1}{5}\right)\) = 6
⇒ a × \(\frac{6}{5}\) = 6 ⇒ a = 5
Thus the required G.P is given by
a, ar, ar², …. i.e. 5, 5 × \(\frac{1}{5}\), 5 × \(\left(\frac{1}{5}\right)^2\), …..
i.e. 5, 1, \(\frac{1}{5}\), …..
∴ S_∞ = \(\frac{a}{1-r}\) = \(\frac{5}{1-\frac{1}{5}}\) = \(\frac{5}{\frac{4}{5}}\) = \(\frac{25}{4}\)

Question 14.
14. Three numbers are in A.P. and their sum is 15 . If 1, 4 and 19 be added to these numbers respectively, the numbers are in G.P. Find the numbers.
Solution:
Let the three numbers in A.P are
a – d, a, a + d.
Since their sum is 15.
∴ a – d + a + a + d = 15
⇒ 3a = 15 ⇒ a = 5
When 1, 4 and 19 be added to these numbers we get the numbers
a – d + 1, a + 4, a + d + 19 are in G.P.
Thus, (a + 4)² = (a – d + 1)(a + d + 19)
⇒ (5 + 4)² = (5 – d + 1)(5 + d + 19)
⇒ 81 = (6 – d)(24 + d)
⇒ 81 = – d² – 18d + 144
⇒ d² + 18d – 63 = 0
∴ d = \(\frac{-18 \pm \sqrt{324+252}}{2}\) = \(\frac{-18 \pm 24}{2}\) = 3, – 21
When a = 5, d = 3
Then required numbers are ;
5 – 3, 5, 5 + 3 i.e. – 2, 5, 8
When a = 5, d = – 21
Then required numbers are;
5 + 21, 5, 5 – 21 i.e. 26, 5, -16

Question 15.
Calculate the least number of terms of the geometric progression 5 + 10 + 20 + …. whose sum would exceed 10,00,000. (ISC)
Solution:
When G.P be 5 + 10 + 20 + ….
With first term a = 5;
common ratio = r = \(\frac{10}{5}\) = 2 > 1
We know that S_n = \(\frac{a\left(r^n-1\right)}{r-1}\)
∴ S_n = \(\frac{5\left(2^n-1\right)}{2-1}\) = 5(2ⁿ – 1)
Now we want to find the least value of n for which S_n > 10,00,000
⇒ 5(2ⁿ – 1) > 1000000
⇒ 2ⁿ – 1 > 200000 ⇒ 2ⁿ > 200001
When n = 17, we have
2¹⁷ = 131072 < 200001
When n = 18, we have
2¹⁸ = 262144 > 200001
Hence the least value of n be 18 .

Question 16.
If S be the sum, P the product and R the sum of the reciprocals of n terms in G.P., prove that P² = \(\left(\frac{\mathrm{S}}{\mathrm{R}}\right)^n\).
Solution:
Let a be the first term and r be the common ratio of given G.P.
S = \(\frac{a\left(r^n-1\right)}{r-1}\)
and P = a . ar . ar² ….. ar^n-1
= aⁿ r^{1+2+3….+n-1}
= aⁿ a\(\frac{n-1}{2}[1+n-1]\) = aⁿ r\(\frac{n(n-1)}{2}\)
and R = \(\frac { 1 }{ a }\) + \(\frac { 1 }{ ar }\) + \(\frac{1}{a r^2}\) + …. n terms

Question 17.
Find the sum of the first n terms of the series : 0.2 + 0.22 + 0.222 + …..
Solution:

Question 18.
If \(\frac { 2 }{ 3 }\) = \(\left(x-\frac{1}{y}\right)\) + \(\left(x^2-\frac{1}{y^2}\right)\) + …. to and xy = 2, then calculate the values of x and y with the condition that x < 1.
Solution:

Question 19.
S₁, S₂, S₃, ……, S_n are sums of n infinite geometric progressions. The first terms of these progressions are 1, 2² – 1, 2³ – 1, ……., 2ⁿ – 1 and the common ratios are \(\frac { 1 }{ 2 }\), \(\frac{1}{2^2}\), \(\frac{1}{2^3}\), ……, \(\frac{1}{2^n}\). Calculate the value of S₁ + S₂ + …… + S_n.
Solution:
Given S₁ be an infinite G.P with first term a₁ = 1 and r₁ = \(\frac { 1 }{ 2 }\) < 1
∴ S₁ = \(\frac{a_1}{1-r_1}\) = \(\frac{1}{1-\frac{1}{2}}\) = 2
S₂ be an infinite G.P with first term
a₂ = 2² – 1 and common ratio r₂ = \(\frac{1}{2^2}\)
∴ S₂ = \(\frac{a_2}{1-r_2}\) = \(\frac{2^2-1}{1-\frac{1}{4}}\) = \(\frac{3}{3}\) = 4
S₃ be an infinite G.P with first term
a₃ = 2³ – 1 and common ratio r₃ = \(\frac{1}{2^3}\)
∴ S₃ = \(\frac{a_3}{1-r_3}\) = \(\frac{2^3-1}{1-\frac{1}{8}}\) = \(\frac{7}{\frac{7}{8}}\) = 8
S_n be an infinite G.P with first term
a_n = 2ⁿ – 1 and common ratio = r_n = \(\frac{1}{2^n}\)
∴ S_n = \(\frac{a_n}{1-r_n}\) = \(\frac{2^n-1}{1-\frac{1}{2^n}}\) = \(\frac{\left(2^n-1\right) 2^n}{2^n-1}\) = 2ⁿ
Then S₁ + S₂ + S₃ + …. + S_n = 2 + 4 + 8 + … + 2ⁿ
which form G.P with a = 2 and r = 2 > 1
= \(\frac{a\left(r^n-1\right)}{r-1}\) = \(\frac{2\left(2^n-1\right)}{2-1}\) = \(2\left(2^n-1\right)\)

Question 20.
Find three numbers a, b, c between 2 and 18 such that :
(i) their sum is 25 , and
(ii) the numbers 2, a, b are consecutive terms of an arithmetic progression, and
(iii) the numbers b, c 18 are consecutive terms of a geometric progression.
Solution:
Given a, b, c are three numbers between 2 and 18 .
it is given that a + b + c = 25 …(1)
Since 2, a, b are consecutive terms of an A.P.
2a = 2 + b ⇒ b = 2a – 2 …(2)
Further b, c, 18 are consecutive terms of G.P.
∴ c² = 18b ⇒ c = \(\sqrt{18 b}\) …(3)
Using eqn. (2) and (3) in eqn. (1); we have
\(\frac{2+b}{2}\) + b + \(\sqrt{18 b}\) = 25
⇒ 2 + b + 2b + 2\(\sqrt{18 b}\) = 50
⇒ 2\(\sqrt{18 b}\) = 48 – 3b ⇒ 6\(\sqrt{2 b}\) = 48 – 3b
⇒ 2\(\sqrt{2 b}\) = 16 – b
On squaring both sides; we have
8b = (16 – b)² ⇒ 8b = b² + 256 – 32b
⇒ b² – 40b + 256 = 0
⇒ b = \(\frac{40 \pm \sqrt{1600-1024}}{2}\)
⇒ b = \(\frac{40 \pm 24}{2}\) = 32, 8
since all the three numbers lies between 2 and 18
∴ b = 8 is acceptable.
∴ from (3); c = \(\sqrt{18 \times 8}\) = \(\sqrt{144}\) = 12
∴ from (1); a = 25 – 8 – 12 = 5
Hence a = 5, b = 8 and c = 12

Question 21.
Three numbers, whose sum is 21 , are in A.P. If 2, 2, 14 are added to them respectively, the resulting numbers are in G.P. Find the numbers.
Solution:
Let the three numbers in A.P are
a – d, a, a + d
and their sum is 21 .
∴ a – d + a + a + d = 21
⇒ 3a = 21 ⇒ a = 7
Further a – d + 2, a + 2, a + d + 14 are in G.P.
∴ (a + 2)² = (a – d + 2)(a + d + 14)
⇒ (7 + 2)² = (7 – d + 2)(7 + d + 14)
⇒ 81 = (9 – d)(21 + d)
⇒ 81 = 189 – 12d – d²
⇒ d² + 12d – 108 = 0
⇒ (d – 6)(d + 18) = 0
⇒ d = 6, -18
When a = 7 and d = 6
Then required numbers are ;
7 – 6, 7, 7 + 6 i.e. 1, 7, 13.
When a = 7 and d = -18
Then required numbers are ;
7 + 18, 7, 7 – 18 i.e. 25, 7, -11.

Question 22.
If x = 1 + a + a² + ….. ∞, a < 1
and y = 1 + b + b² + ….. ∞, b < 1, then prove that
1 + ab + a²b² + …. = \(\frac{x y}{(x+y-1)}\)
Solution:
x = 1 + a + a² + ….. ∞
Clearly form infinite G.P with first term = 1
and common ratio = a < 1
∴ x = \(\frac{1}{1-a}\)
L.H.S. = 1 + ab + a²b² …. ∞
Here a = 1 and C.R = ab < 1 (∵ a < 1, b < 1) = \(\frac{1}{1-ab}\)

Question 23.
If S₁, S₂, S₃, S_p are the sums of infinite geometric series whose first terms are 1 , 2, 3, …., p and whose common ratios are \(\frac{1}{2}\), \(\frac{1}{3}\), ….. \(\frac{1}{p+1}\) respectively, Prove that S₁ + S₂ + S₃ + …….. + S_p = \(\frac{1}{2}\)p (p + 3).
Solution:
S₁ = infinite G.P with first term a₁ = 1
and r₁ = \(\frac{1}{2}\) < 1
∴ S₁ = \(\frac{a_1}{1-r_1}\) = \(\frac{1}{1-\frac{1}{2}}\) = 2
S₂ = infinite G.P with first term a₂ = 2
and r₂ = \(\frac{1}{3}\) < 1
∴ S₂ = \(\frac{a_2}{1-r_2}\) = \(\frac{2}{1-\frac{1}{3}}\) = \(\frac{2}{\frac{2}{3}}\) = 3
S₃ = infinite G.P with first term a₃ = 3
and r₃ = \(\frac{1}{4}\) < 1
∴ S₃ = \(\frac{a_3}{1-r_3}\) = \(\frac{3}{1-\frac{1}{4}}\) = 4 and so on
S_p = infinite G.P with first term a_p = p