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

Option: 1

33


Option: 2

37


Option: 3

43


Option: 4

47


Answers (1)

best_answer

Let \mathrm{A_{1}= a} & common ratio \mathrm{= r}.
\mathrm{a\cdot ar^{2}\cdot ar^{4}\cdot ar^{6}= \frac{1}{1296}\; \; \Rightarrow \; \; a^{4}r^{12}= \frac{1}{1296}\; \Rightarrow a\: ^{2}r^{6}= \frac{1}{36}\: \Rightarrow \; ar^{2}= \frac{1}{6}}
\mathrm{\Rightarrow A_{4}= \frac{1}{6},\quad A_{2}+A_{4}= \frac{7}{36}}
\mathrm{\Rightarrow A_{2}= \frac{7}{36}-\frac{1}{6}= \frac{1}{36}= ar\Rightarrow \: r^{2}= 6\: \Rightarrow \: r= \sqrt{6},a= \frac{1}{36\sqrt{6}}}

\mathrm{A_{6}+A_{8}+A_{10}= ar^{5}+ar^{7}+ar^{9}= \frac{1}{36\sqrt{6}}\left ( 6^{2}\sqrt{6}+6^{3}\sqrt{6} +6^{4}\sqrt{6}\right )}
                                                                      \mathrm{= 1+6+6^{2}= 43}

Option (C)


 

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