If the sum of the first 20 terms of the series

Question:

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

 

  1. $746 / 21$

  2. $7^{1 / 2}$

  3. $\mathrm{e}^{2}$

  4. $7^{2}$


Correct Option: , 4

Solution:

$460=\log _{7} x \cdot(2+3+4+\ldots \ldots+20+21)$

$\Rightarrow 460=\log _{7} x \cdot\left(\frac{21 \times 22}{2}-1\right)$

$\Rightarrow 460=230 \cdot \log _{7} x$

$\Rightarrow \log _{7} x=2 \Rightarrow x=49$

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