Question

Let S_n(x) = `log_a  1/2 x + log_a  1/3 x + log_a  1/6 x + log_a  1/11 x  +  log_a  1/18 x + log_a  1/27x  + ` ...... up to n-terms, where a > 1. If S₂₄(x) = 1093 and S₁₂(2x) = 265, then value of a is equal to ______.

Options

• 15

• 16

• 17

• 18

Answer 1

Let S_n(x) = `log_a  1/2 x + log_a  1/3 x + log_a  1/6 x + log_a  1/11 x  +  log_a  1/18 x + log_a  1/27x  + ` ...... up to n-terms, where a > 1. If S₂₄(x) = 1093 and S₁₂(2x) = 265, then value of a is equal to 16.

Explanation:

S_n = (2 + 3 + 6 + 11 + 18 + 27 + ....)log_a x

Let S = 2 + 3 + 6 + 11  ......  T_n
      S = 2 + 3 + 6 + ......  T_n–1 + T_n
      –   –   –   –    –          –     –
      0 = 2 + 1 + 3 + 5  ...... –T_n 

T_n = `2 + (n - 1)/2[2 + (n - 2)2]` = n² – 2n + 3

∴ S_n = `sum(n^2 - 2n + 3)log_x`

= `((n(n + 1)(2n + 1))/6 - 2(n(n + 1))/2 + 3n)log_ax`

S_n(x) = `n/6[2n^2 - 3n + 13]log_ax`

∴ S₂₄(x) = 1093

⇒ 4 × 1093 log_a x = 1093

log_a x = `1/4`

⇒ x = `a^(1/4)`

⇒ a = x⁴  ...(i)

Now, S₁₂(2x) = 265

⇒ 2(265) log_a 2x = 265

⇒ 2x = `a^(1/2)`

⇒ a = 4x²  ...(ii)

From (i) and (ii)

x⁴ = 4x²

⇒ x² = 4  ...(∵ x ≠ 0)

∴ a = x⁴ = 16