JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Sequences and Series

Let the sum of an infinite G.P., whose first term is a and the common ratio is r , be 5 . Let the sum of its first five terms be

\frac{98}{25}

. Then the sum of the first 21 terms of an AP, whose first term is

10 ar, n^{th}

term is

a_{n}

and the common difference is

10 ar^{2}

, is equal to :

Answer: (A)

Solution

From the infinite G.P.:

\frac{a}{1 - r} = 5

, so

a = 5 (1 - r)

.

From the sum of first 5 terms:

\frac{a ( 1 - r ^{5} )}{1 - r} = \frac{98}{25}

, which gives

a (1 - r^{5}) = \frac{98 ( 1 - r )}{25}

.

For the A.P. with first term

a_{1} = 10 a r

and common difference

d = 10 a r^{2}

S_{21} = \frac{21}{2} [2 a_{1} + 20 d] = \frac{21}{2} [20 a r + 200 a r^{2}] = 21 [10 a r + 100 a r^{2}]

The 11th term is

a_{11} = 10 a r + 10 \cdot 10 a r^{2} = 10 a r + 100 a r^{2}

.

Therefore,

S_{21} = 21 \cdot a_{11}

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Single correct option