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JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Sequences and Series

Let

S_{n} (x) = lo g_{a^{1/2}} x + lo g_{a^{1/3}} x + lo g_{a^{1/6}} x + lo g_{a^{1/11}} x + lo g_{a^{1/18}} x + lo g_{a^{1/27}} x + \dots

. up to

n

-terms. Where

a >

1. If

S_{24} (x) = 1093

and

S_{12} (2 x) = 265

, then the value of

a

is equal to.....

Answer: 16

Solution

Since

lo g_{a^{1/ k}} x = k lo g_{a} x

, the

m

-th term contributes a factor

k_{m}

, and the numbers

2, 3, 6, 11, 18, 27, \dots

increase by the odd numbers

1, 3, 5, \dots

, so

k_{m} = 2 + (m - 1)^{2}

. Hence

S_{n} (x) = [2 n + j = 0 \sum n - 1 j^{2}] lo g_{a} x = [2 n + \frac{( n - 1 ) n ( 2 n - 1 )}{6}] lo g_{a} x

For

n = 24

the bracket is

48 + 4324 = 4372

, so

4372 lo g_{a} x = 1093

and

lo g_{a} x = \frac{1}{4}

. For

n = 12

the bracket is

24 + 506 = 530

, so

530 lo g_{a} (2 x) = 265

and

lo g_{a} (2 x) = \frac{1}{2}

. Subtracting,

lo g_{a} 2 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \Rightarrow a^{1/4} = 2

so

a = 16

.

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Integer answer

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