RBSE Solutions for Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.5

Rajasthan Board RBSE Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.5

Question 1.
Find the sum of n terms of following series:

Solution:

Let sum of this series is S_n

Hence,arithmetic geometric series

(ii) 1 + 3x + 5x² + 7x³ + …
In the given series 1, 3, 5, 7, … are in A.P. and 1, x, x², … are in GP.
The nth term of A.P. is (2n – 1) and nth term of G.P. is x^{n – 1} .
Hence, nth term of arithmetic-geometric series will be (2n – 1)x^{n – 1} .
Let sum of n terms of arithmetic-geometric series is S_n then
S_n = 1 + 3x + 5x² + 7x³ + … (2n – 1)x^{n – 1} … (i)
Multiplying on both sides by x
xS_n = x + 5x² + 3x³ + … [2(n – 1) – 1 ]x^{n – 1} 5(2n – 1)xⁿ … (ii)
Subtracting equation (ii) from (i), we get
(1 – x)S_n = (1 + 2x + 2x² + 2x³ + … + 2x^{n – 1}) + (2n – 1)xⁿ

Hence, it is sum of n terms.

Let sum of its n terms is S_n then

Subtracting equation (ii) from (i), we get

Question 2.
Find the sum of infinite terms of following series :

Solution:
(i) Let sum of the given series is S_∞ then

Subtracting equation (ii) from (i), we get

Hence, sum of the series is 6/7.
(ii) Let sum of the given series is S_∞ then

Adding equation (i) and (ii), we get

Hence, sum of the series is 3/16.
(iii) Let sum of the given series is S_∞ then
S_∞ = 2x + 3x² – 4x³
S_∞ = x – 2x² + 3x³
Adding equation (i) and (ii), we get
(1 + x)S_∞ x – 2x ² 3x ³ +…

Question 3.
Find n^th term and sum of n terms of following series :
(i )2 + 5 + 14 + 41 + 122 + …
(ii) 3.2 + 5.2² + 7.2³ + …
(iii) 1 + 4x + 7x² + 10x³ + …
Solution:
(i) Difference of consecutive terms of given series 3,9, 27,… are in G.P.
Let sum of its n terms is S_n and n^th term is T_n , then
S_n  =2 + 5 + 14 + 41 + 122 + … + T_n … (i)
By increasing on place
S_n  =2 + 5+ 14 + 41 + 122 + … + T_{n – 1} + T_n … (ii)
Subtracting equation (ii) from (i), we get
0 = 2 + (3 + 9 + 27 + … + (n – 1) term) – T_n
⇒ T_n = 2 + (3 + 9 + 27 +… + (n- 1) term)

Let the sum of series is S_n on putting the value of n = 1, 2, 3, …, then

(ii) In the given series, A.P. is 3, 5, 7, … and its n ^th term
= 3 + (n – 1)2 = 3 + 2n – 2 = 2n + 1. and G.P.
is 2, 2², 2³, … and its n^th term is 2ⁿ.
Let sum of its n terms is S_n and n^th term is T_n ,
Then term of given arithmetic-geometric series is
T_n = (2n + 1)2ⁿ.
Let the sum of series is S_n on putting the value of n in this, then
S_n = 3.2 + 5.2² + 7.2³ + … +(2n – 1)2^{n + 1} + (2n+ 1)2ⁿ …(i)
2S_n = 3.2² + 5.2³ +…+ (2n – 1)2ⁿ + (2n + 1)2^{n + 1}…(ii)
Subtracting equation (ii) from (i), we get
-S_n = 3.2 + 2[2² + 2³ + …2ⁿ] -(2n+ 1)2^{n + 1}

⇒ -S_n = 6 + 2² (2^{n + 1} – 1) – (2n + 1)2^n + 1
⇒ -S_n = 6 + 2^{n + 2} – 8 – (2n + 1)2^{n + 1}
⇒ -S_n = 2ⁿ⁺¹ [2 – 2n – 1] – 2
⇒ -S_n = (2n- 1)2^{n + 1} + 2
Hence, sum of n terms
= (2n + 5)2^{n + 1} + 2.

(iii) In the given series, A.P. is 1, 4, 7, 10,… and its n^th term
= 1 + (n – 1)³ = 3n – 2 and G.P.
is 1,x, x², x³,… and its n^th term = x^{n – 1}
Hence, n^th term of arithmetic-geometric series is (3n – 2)x^{n – 1}and sum is S_n
than S_n= 1 + 4x + 7x² + 10x³ +…+ (3n – 5)x^{n – 2}
+ (3n – 2)x^{n – 1} … (i)
xS_n = x + 4x² + 7x³ +…+ (3n – 5)x^{n – 1}
+ (3n – 2)x^{n – 1} … (ii)
Subtracting equation (ii) from (i), we get
(1 – x)S_n = 1 + [3x + 3x² + 3x³ +. ..3x^{n – 1}] – (3n- 2)x^{n + 1}

Question 4.
Find the sum of n terms of series 2 + 5x + 8x² + 11x³ + … and hence obtained the sum of infinite series | x | < 1.
Solution:
In the given series, A. P. is 2, 5, 8, 11, … and its n^th term
= 2 + (n – 1 )3 = 2 + 3n – 3 = 3n – 1 and G.P
is 1, x, x², x³,… and its nth term is x^{n – 1} and sum of
arithmetic-geometric series be S_n and its term is (2n – 1)x^{n – 1} so
S_n = 2 + 5x + 8x² + 11x³ +…+ (3n – 4)x^{n – 2} + (3n – 1)x^{n – 1} … (i)
xS_n = 2x + 5x² + 8x³ + … + (3n – 4)x^{n –}  + (3n- 1)xⁿ … (ii)
Subtracting equation (ii) from (i), we get
(1 – x)S_n = 2 + [3x + 3x² + 3x³ + … + 3x^{n – 1}] + (3n – 1)xⁿ

RBSE Solutions for Class 11 Maths