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JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Sequences and Series

If sum of the first 21 terms of the series

lo g_{9^{1/2}} X + lo g_{9^{1/3}} X + lo g_{9^{1/4}} X + \dots

, where

x > 0

is 504 , then

x

is equal to :

(A)7
(B)9
(C)243
(D)81✓ Correct

Answer: (D)

Solution

The

k

-th term of the series has base

9^{1/ (k + 1)}

, and

lo g_{9^{1/ (k + 1)}} x = (k + 1) lo g_{9} x

Summing over

k = 1

to

21

multiplies

lo g_{9} x

by

2 + 3 + \dots + 22 = \frac{22 \times 23}{2} - 1 = 252

. Hence

252 lo g_{9} x = 504

, so

lo g_{9} x = 2

and

x = 9^{2} = 81

.

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Single correct option

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