If the sum of the first 20 terms of the
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. (1) $7^{2}$

2. (2) $7^{1 / 2}$

3. (3) $e^{2}$

4. (4) $7^{46 / 21}$

Correct Option: 1

Solution:

$S=\log _{7} x^{2}+\log _{7} x^{3}+\log _{7} x^{4}+\ldots 20$ terms

$\because S=460$

$\Rightarrow \log _{7}\left(x^{2} \cdot x^{3} \cdot x^{4} \cdot \ldots \ldots x^{21}\right)=460$

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

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

$\Rightarrow \frac{20}{2}(2+21) \log _{7} x=460$

$\Rightarrow \log _{7} x=\frac{460}{230}=2 \Rightarrow x=7^{2}=49$