Prove the following
Question:

Let

$S_{n}(x)=\log _{a^{1 / 2}} x+\log _{a^{1 / 3}} x+\log _{a^{1 / 6}} x$ $+\log _{a^{1 / 11}} x+\log _{a^{1 / 18}} x+\log _{a^{1 / 27}} x+\ldots \ldots$

up to $\mathrm{n}$-terms, where $\mathrm{a}>1$. If $\mathrm{S}_{24}(\mathrm{x})=1093$ and $S_{12}(2 x)=265$, then value of a is equal to_____.

Solution:

$S_{n}(x)=(2+3+6+11+18+27+\ldots \ldots+n-$ terms $) \log _{a} x$

Let $\mathrm{S}_{1}=2+3+6+11+18+27+\ldots .+\mathrm{T}_{\mathrm{n}}$

$\mathrm{S}_{1}=2+3+6+\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots+\mathrm{T}_{\mathrm{n}}$

_____________________

$\mathrm{T}_{\mathrm{n}}=2+1+3+5+\ldots \ldots+\mathrm{n}$ terms

$\mathrm{T}_{\mathrm{n}}=2+(\mathrm{n}-1)^{2}$

$S_{1}=\Sigma T_{n}=2 n+\frac{(n-1) n(2 n-1)}{6}$

$\Rightarrow S_{n}(x)=\left(2 n+\frac{n(n-1)(2 n-1)}{6}\right) \log _{a} x$

$\mathrm{S}_{24}(\mathrm{x})=1093$ (Given)

$\log _{\mathrm{a}} \mathrm{x}\left(48+\frac{23.24 .47}{6}\right)=1093$

$\log _{\mathrm{a}} \mathrm{x}=\frac{1}{4}$…..(1)

$S_{12}(2 x)=265$

$S_{12}(2 x)=265$

$\log _{\mathrm{a}}(2 \mathrm{x})\left(24+\frac{11.12 .23}{6}\right)=265$

$\log _{\mathrm{a}} 2 \mathrm{x}=\frac{1}{2}$….(2)

$(2)-(1)$

$\log _{\mathrm{a}} 2 \mathrm{x}-\log _{\mathrm{a}} \mathrm{x}=\frac{1}{4}$

$\log _{\mathrm{a}} 2=\frac{1}{4} \Rightarrow \mathrm{a}=16$