Question

If the sum of the first $20$ terms of the series ${\log }_{\left(7^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}x+{\log }_{\left(7^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}x+{\log }_{\left(7^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\right)}x+...........$ is $460$, then $x$ is equal to:

• A. $7^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$
• B. $7^2$
• C. $e^2$
• D. $7^{\raisebox{1ex}{$46$}\!\left/ \!\raisebox{-1ex}{$21$}\right.}$

Answer 1

The correct option is B

$7^2$

Explanation for the correct option.

$$\begin{gathered} {\log }_{\left(7^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}x+{\log }_{\left(7^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}x+{\log }_{\left(7^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\right)}x+...........20terms \\ =\frac{1}{{\displaystyle \frac{1}{2}}}{\log }_{7}x+\frac{1}{{\displaystyle \frac{1}{3}}}{\log }_{7}x+\frac{1}{{\displaystyle \frac{1}{4}}}{\log }_{7}x+...........20terms\left[By{\log }_{a^n}{b}^m=\frac{m}{n}·{\log }_{a}b\right] \\ =2{\log }_{7}x+3{\log }_{7}x+4{\log }_{7}x+...........20terms \\ ={\log }_{7}x\left(2+3+4+...........20terms\right) \end{gathered}$$

Here, $2+3+4+.....$are in AP where, $a=2,d=1,\mathrm{and}n=20$

$\Rightarrow \left(2+3+4+...........20terms\right)=S_{20}$

$\begin{array}{rcl}{S}_{20}& =& \frac{20}{2}\left[2\left(2\right)+19\left(1\right)\right]\left[By{S}_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\right]\\ & =& 10\left(23\right)\\ & =& 230\end{array}$

$\begin{array}{rcl}\Rightarrow {\log }_{7}x\left(2+3+4+...........20terms\right)& =& {\log }_{7}x\left(230\right)\\ \Rightarrow 460& =& {\log }_{7}x\left(230\right)\left(given\right)\\ \Rightarrow 2& =& {\log }_{7}x\\ \Rightarrow x& =& 7^2\end{array}$

Hence, option B is correct.