If the sum of the first 21 terms of the serie | vedclass

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(A) The given series is $\log _{9^{1/2}} x + \log _{9^{1/3}} x + \log _{9^{1/4}} x + \dots$Using the property $\log_{a^b} x = \frac{1}{b} \log_a x$,th

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$81$

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(A) The given series is $\log _{9^{1/2}} x + \log _{9^{1/3}} x + \log _{9^{1/4}} x + \dots$Using the property $\log_{a^b} x = \frac{1}{b} \log_a x$,the terms become:$2 \log_9 x + 3 \log_9 x + 4 \log_9 x + \dots$This is an arithmetic series with $21$ terms where the first term $a = 2 \log_9 x$ and the common difference $d = \log_9 x$.The sum of $n$ terms is $S_n = \frac{n}{2} [2a + (n-1)d]$.For $n = 21$,$S_{21} = \frac{21}{2} [2(2 \log_9 x) + (21-1) \log_9 x] = 504$.$S_{21} = \frac{21}{2} [4 \log_9 x + 20 \log_9 x] = \frac{21}{2} [24 \log_9 x] = 21 	imes 12 \log_9 x = 252 \log_9 x$.Given $252 \log_9 x = 504$,we get $\log_9 x = 2$.Therefore,$x = 9^2 = 81$.

If the sum of the first $21$ terms of the series $\log _{9^{1 / 2}} x + \log _{9^{1 / 3}} x + \log _{9^{1 / 4}} x + \dots$ where $x > 0$ is $504$,then $x$ is equal to:

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