Let S_{n}(x) = log_{a^{1/2}} x + log_{a^{1/3} | vedclass

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(A) The general term of the series is $\log_{a^{1/k_n}} x = k_n \log_a x$,where $k_n$ follows the sequence $2, 3, 6, 11, 18, 27, \ldots$.The differenc

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$16$

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(A) The general term of the series is $\log_{a^{1/k_n}} x = k_n \log_a x$,where $k_n$ follows the sequence $2, 3, 6, 11, 18, 27, \ldots$.The differences between consecutive terms are $1, 3, 5, 7, 9, \ldots$,which is an arithmetic progression.The $n$-th term of this sequence is $k_n = 2 + \sum_{i=0}^{n-1} (2i-1)$ for $n > 1$,which simplifies to $k_n = (n-1)^2 + 2$.The sum of the first $n$ terms is $S_n(x) = \left( \sum_{i=1}^n ((i-1)^2 + 2) ight) \log_a x = \left( \frac{(n-1)n(2n-1)}{6} + 2n ight) \log_a x$.For $n=24$,$S_{24}(x) = \left( \frac{23 	imes 24 	imes 47}{6} + 48 ight) \log_a x = (4324 + 48) \log_a x = 4372 \log_a x = 1093$,so $\log_a x = \frac{1093}{4372} = \frac{1}{4}$.For $n=12$,$S_{12}(2x) = \left( \frac{11 	imes 12 	imes 23}{6} + 24 ight) \log_a (2x) = (506 + 24) \log_a (2x) = 530 \log_a (2x) = 265$,so $\log_a (2x) = \frac{265}{530} = \frac{1}{2}$.Subtracting the two equations: $\log_a (2x) - \log_a x = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$.$\log_a 2 = \frac{1}{4} \implies a^{1/4} = 2 \implies a = 2^4 = 16$.

Let $S_{n}(x) = \log_{a^{1/2}} x + \log_{a^{1/3}} x + \log_{a^{1/6}} x + \log_{a^{1/11}} x + \log_{a^{1/18}} x + \log_{a^{1/27}} x + \ldots$ up to $n$-terms,where $a > 1$. If $S_{24}(x) = 1093$ and $S_{12}(2x) = 265$,then the value of $a$ is equal to ..... .

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