JEE Main 2021 (Online) 16th March Evening ShiftQuestion 1
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 $S_{12}(2x)=265$, then value of $a$ is equal to $\_\_\_\_\_\_$ .
Sequence and Series · Standard Sums (Σ formulas)
Answer$16$
Solution
Formulas used. (1) $\log_{a^{1/k}}x=k\log_ax$(2) $\sum_{j=1}^{n}(j-1)^2=\frac{(n-1)n(2n-1)}{6}$Step 1 - The multipliers. $2,3,6,11,18,27,\dots$ have differences $1,3,5,7,\dots$, so $k_j=(j-1)^2+2$. Step 2 - $S_n$. $$S_n(x)=\left[\frac{(n-1)n(2n-1)}{6}+2n\right]\log_ax .$$ $n=24$: $4324+48=4372$. $\quad n=12$: $506+24=530$. Step 3 - Use the data. $$4372\log_ax=1093\ \Longrightarrow\ \log_ax=\frac14,$$ $$530\log_a(2x)=265\ \Longrightarrow\ \log_a2+\log_ax=\frac12\ \Longrightarrow\ \log_a2=\frac14 .$$ Step 4. $a^{1/4}=2$, so $a=16$. Answer: 16