If the sum of the first 20 terms of the serie | vedclass

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(D) The given series is $\log_{(7^{1/2})} x + \log_{(7^{1/3})} x + \log_{(7^{1/4})} x + \dots + \log_{(7^{1/21})} x = 460$.Using the property $\log_{(

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$7^2$

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(D) The given series is $\log_{(7^{1/2})} x + \log_{(7^{1/3})} x + \log_{(7^{1/4})} x + \dots + \log_{(7^{1/21})} x = 460$.Using the property $\log_{(a^n)} x = \frac{1}{n} \log_a x$,we can rewrite the terms as:$2 \log_7 x + 3 \log_7 x + 4 \log_7 x + \dots + 21 \log_7 x = 460$.Factoring out $\log_7 x$,we get:$\log_7 x \cdot (2 + 3 + 4 + \dots + 21) = 460$.The sum of the arithmetic progression $2 + 3 + \dots + 21$ is given by $\frac{n}{2}(a + l)$,where $n = 20$,$a = 2$,and $l = 21$.Sum $= \frac{20}{2}(2 + 21) = 10 	imes 23 = 230$.So,$230 \cdot \log_7 x = 460$.$\log_7 x = \frac{460}{230} = 2$.Therefore,$x = 7^2 = 49$.

If the sum of the first $20$ terms of the series $\log_{(7^{1/2})} x + \log_{(7^{1/3})} x + \log_{(7^{1/4})} x + \dots$ is $460$,then $x$ is equal to

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