Let A _{1}, A _{2}, A _{3}, ldots ldots be an increasing geometric progression of positive real numb

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8. Sequences and Series

hard

Let $A _{1}, A _{2}, A _{3}, \ldots \ldots$ 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

A

$33$

B

$37$

C

$43$

D

$47$

(JEE MAIN-2022)

Solution

$A _{1} \cdot A _{3} \cdot A _{5} \cdot A _{7}=\frac{1}{1296}$

$\left( A _{4}\right)^{4}=\frac{1}{1296}$

$A _{4}=\frac{1}{6}…..(1)$

$A _{2}+ A _{4}=\frac{7}{36}$

$A _{2}=\frac{1}{36}…..(2)$

$A _{6}=1$

$A _{8}=6$

$A _{10}=36$

$A _{6}+ A _{8}+ A _{10}=43$

Standard 11

Mathematics

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