Question

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

Answer 1

$S_n\left(x\right)={\log }_{a^{1/2}}x+{\log }_{a^{1/3}}x+{\log }_{a^{1/6}}x+{\log }_{a^{1/11}}x+{\log }_{a^{1/18}}x+{\log }_{a^{1/27}}x+\cdots \text{up to}n\text{-terms}$
$\Rightarrow S_n\left(x\right)=2{\log }_{a}x+3{\log }_{a}x+6{\log }_{a}x+11{\log }_{a}x+\cdots$
$\Rightarrow S_n\left(x\right)=\left({\log }_{a}x\right)(2+3+6+11+\cdots )$

$S_r=2+3+6+11+\cdots$
General term, $T_r=r^2-2r+3$
$S_n\left(x\right)=\sum _{r=1}^n\left({\log }_{a}x\right)(r^2-2r+3)$
$\Rightarrow S_{24}\left(x\right)=\left({\log }_{a}x\right)\sum _{r=1}^{24}(r^2-2r+3)$
$\Rightarrow 1093=4372{\log }_{a}x$
$\Rightarrow {\log }_{a}x=\frac{1}{4}$
$\Rightarrow x=a^{1/4}\cdots \left(1\right)$

$S_{12}\left(2x\right)={\log }_a\left(2x\right)\sum _{r=1}^{12}(r^2-2r+3)$
$\Rightarrow 265=530{\log }_a\left(2x\right)$
$\Rightarrow {\log }_a\left(2x\right)=\frac{1}{2}$
$\Rightarrow 2x=a^{1/2}\cdots \left(2\right)$

After solving $\left(1\right)$ and $\left(2\right),$ we get
$a^{1/4}=2$
$\Rightarrow a=16$