Question

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

Answer 1

Answer: (C)

$A _{1} \cdot A _{3} \cdot A _{5} \cdot A _{7}=\frac{1}{1296}$
$\left( A _{4}\right)^{4}=\frac{1}{1296}$
$A _{4}=\frac{1}{6}$ ...(1)
$A _{2}+ A _{4}=\frac{7}{36}$
$A _{2}=\frac{1}{36}$ ...(2)
$A _{6}=1$
$A _{8}=6$
$A _{10}=36$
$A _{6}+ A _{8}+ A _{10}=43$