Question

Let
$S_{n}(x)= \log _{a^{ 1 / 2}} x+\log _{a^{1 / 3}} x+\log _{a^{1 / 6}} x $
$+\log _{a^{1 / 11}} x+\log _{a^{1 / 18}} x+\log _{a^{1 / 27}} x+\ldots .$
up to n-terms, where $a >1$. If $S _{24}( x )=1093$ and $S _{12}(2 x )=265$, then value of a is equal to _______.

Answer 1

$S _{ n }( x )=(2+3+6+11+18+27+\ldots \ldots+ n$ -terms $) \log _{ a } x$
Let $S _{1}=2+3+6+11+18+27+\ldots+ T _{ n }$
$S _{1}=2+3+6+\ldots \ldots \ldots \ldots \ldots .+ T _{ n }$
$T_{n}=2+1+3+5+\ldots \ldots \ldots + n$ terms
${ T _{ n }}=2+( n -1)^{2}$
$S _{1}=\Sigma T _{ n }=2 n +\frac{( n -1) n (2 n -1)}{6}$
$\Rightarrow S _{ n }( x )=\left(2 n +\frac{ n ( n -1)(2 n -1)}{6}\right) \log _{ a } x$
$S _{24}( x )=1093$ (Given)
$\log _{ a } x \left(48+\frac{23.24 .47}{6}\right)=1093$
$\log _{ a } x =\frac{1}{4} \ldots .(1)$
$S _{12}(2 x )=265$
$S _{12}(2 x )=265$
$\log _{ a }(2 x )\left(24+\frac{11.12 .23}{6}\right)=265$
$\log _{ a } 2 x =\frac{1}{2} \ldots .(2)$
$(2)-(1)$
$\log _{ a } 2 x -\log _{ a } x =\frac{1}{4}$
$\log _{ a } 2=\frac{1}{4} \Rightarrow a =16$