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(C) Let the geometric progression be $a, ar, ar^2, \ldots$ where $a > 0$ and $r > 1$.Given $A_{1} A_{3} A_{5} A_{7} = \frac{1}{1296}$.Substituting the

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$43$

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(C) Let the geometric progression be $a, ar, ar^2, \ldots$ where $a > 0$ and $r > 1$.Given $A_{1} A_{3} A_{5} A_{7} = \frac{1}{1296}$.Substituting the terms: $a \cdot (ar^2) \cdot (ar^4) \cdot (ar^6) = a^4 r^{12} = (ar^3)^4 = (A_{4})^4 = \frac{1}{1296}$.Thus,$A_{4} = \sqrt[4]{\frac{1}{1296}} = \frac{1}{6}$.Given $A_{2} + A_{4} = \frac{7}{36}$,we have $ar + ar^3 = \frac{7}{36}$.Since $A_{4} = ar^3 = \frac{1}{6}$,we have $ar + \frac{1}{6} = \frac{7}{36}$,so $ar = \frac{7}{36} - \frac{6}{36} = \frac{1}{36}$.Now,$r^2 = \frac{ar^3}{ar} = \frac{1/6}{1/36} = 6$,so $r = \sqrt{6}$.Then $a = \frac{ar}{r} = \frac{1/36}{\sqrt{6}} = \frac{1}{36\sqrt{6}}$.We need to find $A_{6} + A_{8} + A_{10} = ar^5 + ar^7 + ar^9 = ar^5(1 + r^2 + r^4)$.$ar^5 = (ar) \cdot r^4 = \frac{1}{36} \cdot (6)^2 = \frac{36}{36} = 1$.$A_{6} + A_{8} + A_{10} = 1 \cdot (1 + 6 + 36) = 43$.

Let $A_{1}, A_{2}, A_{3}, \ldots$ be an increasing geometric progression of positive real numbers. If $A_{1} A_{3} A_{5} A_{7} = \frac{1}{1296}$ and $A_{2} + A_{4} = \frac{7}{36}$,then the value of $A_{6} + A_{8} + A_{10}$ is equal to

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