## 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 + 5x^{2} + 7x^{3} + …

In the given series 1, 3, 5, 7, … are in A.P. and 1, x, x^{2}, … 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 + 5x^{2} + 7x^{3} + … (2n – 1)x^{n – 1} … (i)

Multiplying on both sides by x

xS_{n} = x + 5x^{2} + 3x^{3} + … [2(n – 1) – 1 ]x^{n – 1 }5(2n – 1)x^{n} … (ii)

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

(1 – x)S_{n} = (1 + 2x + 2x^{2} + 2x^{3} + … + 2x^{n – 1}) + (2n – 1)x^{n}

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 + 3x^{2} – 4x^{3
}S_{∞ }= x – 2x^{2} + 3x^{3
}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 + 5.2^{2} + 7.2^{3} + …

(iii) 1 + 4x + 7x^{2} + 10x^{3} + …

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, 2^{2}, 2^{3}, … and its n^{th} term is 2^{n}.

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)2^{n}.

Let the sum of series is S_{n} on putting the value of n in this, then

S_{n} = 3.2 + 5.2^{2} + 7.2^{3} + … +(2n – 1)2^{n + 1 }+ (2n+ 1)2^{n} …(i)

2S_{n} = 3.2^{2} + 5.2^{3} +…+ (2n – 1)2^{n} + (2n + 1)2^{n} ^{+ 1}…(ii)

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

-S_{n} = 3.2 + 2[2^{2} + 2^{3} + …2^{n}] -(2n+ 1)2^{n }^{+ 1
}

⇒ -S_{n} = 6 + 2^{2} (2^{n} ^{+ 1} – 1) – (2n + 1)2^{n + }^{1
}⇒ -S_{n} = 6 + 2^{n} ^{+ 2} – 8 – (2n + 1)2^{n} ^{+ 1
}⇒ -S_{n} = 2^{n}^{+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, x^{2}, x^{3},… 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 + 7x^{2 }+ 10x^{3} +…+ (3n – 5)x^{n – 2
}+ (3n – 2)x^{n – 1} … (i)

xS_{n} = x + 4x^{2} + 7x^{3} +…+ (3n – 5)x^{n – 1}

+ (3n – 2)x^{n – 1} … (ii)

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

(1 – x)S_{n} = 1 + [3x + 3x^{2} + 3x^{3} +. ..3x^{n – 1}] – (3n- 2)x^{n + 1} ^{
}

Question 4.

Find the sum of n terms of series 2 + 5x + 8x^{2 }+ 11x^{3} + … 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, x^{2}, x^{3},… 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 + 8x^{2} + 11x^{3} +…+ (3n – 4)x^{n – 2 }+ (3n – 1)x^{n – 1} … (i)

xS_{n} = 2x + 5x^{2} + 8x^{3} + … + (3n – 4)x^{n – }+ (3n- 1)x^{n} … (ii)

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

(1 – x)S_{n} = 2 + [3x + 3x^{2} + 3x^{3} + … + 3x^{n – 1}] + (3n – 1)x^{n
}