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 :
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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