RBSE Solutions for Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.4

Rajasthan Board RBSE Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.4

Question 1.
Find the sum of the following geometric progression :

Solution:
(i) Given progression
= 2 + 6 +18 + 51 +….

Question 2.
Find the sum of following geometric progression :
(i) 2 + 6 + 18 + 54 + … + 486
(ii) 64 + 32 + 16 + … + 1/4
Solution:
(i) Given progression
= 2 + 6 + 18 + 54….+ 486
a = 2, r = 6/2 = 3, T_n = 486 (r > 1)
n^th term T_n = ar^{n – 1}
ar^{n – 1} = 486
2.(3)^{n – 1} = 486
⇒ (3)^{n – 1} = 243 = (3)⁵
⇒ n – 1 = 5
n = 5 + 1 =6
Sum of n terms

s₆ = 3⁶ – 1
= 729 -1 = 728
Hence, sum is 728.
(ii) Given progression

Question 3.
Sum of how many terms of G.P. 4, 12, 36, is 484 ?
Solution:
a = 4, r = 12/4 = 3, S_n = 484, (r >1)
S_n = 484

3ⁿ – 1 = 242
3ⁿ = 242 + 1 = 243
⇒  3ⁿ = 3⁵
Hence, number of term is 5.

Question  4.
Sum of first 5 terms of any G.P. is 124 and common ratio is 2. Find the first term of the series.
Solution:
n = 5, S_n = 124, r = 2, r > 1
S_n = 124

Hence, first term is 4.

Question 5.
The common ratio of any G.P. is 2, last term 160 and sum is 310. Find the first term of the series.
Solution:
r = 2,T_n= 160, S_n = 310, r > 1,
ar^{n – 1} = 160
⇒ a(2)^n – 1 = 160

⇒ a(2)ⁿ – a = 310    … (ii)
From equation (i) and … (ii)
a = 320 – 310 = 10
Hence, first term is 10.

Question 6.
Find the sum of first n terms of the following series :
(i) 7 + 11 + 777 + …
(ii) .5 + .55 + .555 + ….
(iii) .9 + .99 + .999 + ….
Solution:
(i) Given series upto n terms 7+ 77 + 777 + 7777  upto n terms. It is clear that this series is not in G.P. but it can be related to G.P. by writing it in the following form.
S_n = 7 + 77 + 777 + 7777 + … upto n terms
= 7[ 1 + 11 + 111 + 1111 + … upto n terms]

(ii) It is clear that this series is not in G.P. but it can be related to G.P. by writing it in the following form :
S_n =.5 + .55 + .555 + .5555 + ….upto n terms]

Question 7.
Find the rational form of the following recurring decimals :
(i) 2.3\(\bar { 5 } \)
(ii) .6\(\bar { 25 } \)
(iii) 2.\(\bar { 752 } \)
Solution:
(i) Digit 5 is repeated in 2.35 .
∴ 2.35 = 2.35555…………
= 2 + [.3 + .05 + .005 + .0005+ … upto infinity]

Question 8.
First term of any infinite series is 64 and each term is three times its succeeding terms. Find the series.
Solution:
Let infinte series is
a + ar + ar² + ar^-1 + …
Given, First term = 64
According to question,
a = 3 (ar + ar² + ar³ + … ∞)
⇒ a = 3ar(1 + r + r² + … ∞)
⇒ 1 = 3r \(\left( \frac { 1 }{ 1-r } \right) \).
⇒ 1 – r = 3r
⇒ 4r = 1 or r = 1/4

Question 9.
If y = x + x² + x³ + … ∞, where | x | < 1, then Prove that
x=\(\left( \frac { y }{ 1+y } \right) \)
Solution:
y = x + x² + x³ + … ∞
⇒ y = x( 1 + x + x² + … ∞)
⇒ y = x \(\left( \frac { 1 }{ 1-x } \right) \)
⇒ y (1 – x) = x
⇒ y – xy – x =0
⇒ y – x(y + 1) =0
⇒ x(1 + y) = y
⇒ x = \(\left( \frac { y }{ 1+y } \right) \)

Question 10.
If x = 1 + a + a² + … ∞, where | a | < 1 and y = 1 + b + b² + … = ∞, where | b | < 1, then prove that

Solution:
Given x = 1 + a + a² + … ∞ (| a | < 1)

Question 11.
Find the sum of the series

Solution:
Given series

RBSE Solutions for Class 11 Maths