JEE Main 2021 — Sequences and Series Question with Solution

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

$\Rightarrow S_n\left(x\right)=(2+3+6+11+18+27+\dots +n$-terms $){\log }_{a}x$

Let $S_1=2+3+6+11+18+27+\dots +T_n$

$S_1=2+3+6+11+18+...+T_n ...\left(i\right)$

$S_1=0+2+3+6+11+18+...+T_{n-1}+T_n ...\left(ii\right)$

Subtract $\left(i\right) \text{from} \left(ii\right)$we get, 

$T_n=2+1+3+5+...+\text{upto }\left(n-1\right)\text{-term}$

$T_n=2+(n-1{)}^2$

$S_1=\Sigma T_n=2n+\frac{(n-1)n(2n-1)}{6}$

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

$S_{24}\left(x\right)=1093$ (Given)

${\log }_{a}x\left(48+\frac{23.24.47}{6}\right)=1093$

${\log }_{a}x=\frac{1}{4}$ ......$\left(1\right)$

$S_{12}\left(2x\right)=265$

$S_{12}\left(2x\right)=265$

${\log }_a\left(2x\right)\left(24+\frac{11.12.23}{6}\right)=265$

${\log }_{a}2x=\frac{1}{2}$ ......$\left(2\right)$

$\left(2\right)-\left(1\right)$

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

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