Question

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

Answer 1

Answer: 16

$S_n(x)=(2+3+6+11+18+27+\dots \dots +n$ -terms $){\log }_{a}x$
Let $S_1=2+3+6+11+18+27+\dots +T_n$
$S_1=2+3+6+\dots \dots \dots \dots \dots .+T_n$
$T_n=2+1+3+5+\dots \dots \dots +n$ terms
$T_n=2+(n-1{)}^2$
$S_1=\Sigma T_n=2n+\frac{(n-1)n(2n-1)}{6}$
$\Rightarrow S_n(x)=\left(2n+\frac{n(n-1)(2n-1)}{6}\right){\log }_{a}x$
$S_{24}(x)=1093$ (Given)
${\log }_{a}x\left(48+\frac{\mathrm{23.24.47}}{6}\right)=1093$
${\log }_{a}x=\frac{1}{4}\dots .(1)$
$S_{12}(2x)=265$
$S_{12}(2x)=265$
${\log }_a(2x)\left(24+\frac{\mathrm{11.12.23}}{6}\right)=265$
${\log }_{a}2x=\frac{1}{2}\dots .(2)$
$(2)-(1)$
${\log }_{a}2x-{\log }_{a}x=\frac{1}{4}$
${\log }_{a}2=\frac{1}{4}\Rightarrow a=16$