Question
Let $a_{1}, a_{2}, a_{3}, \ldots$ be an $A.P.$ If $\sum_{r=1}^{\infty} \frac{a_{r}}{2^{r}}=4$, then $4 a_{2}$ is equal to

Answer

b
$S=\frac{a_{1}}{2}+\frac{a_{2}}{2^{2}}+\frac{a_{3}}{2^{3}}+\ldots$

$\frac{S}{2} =\frac{a_{1}}{2^{2}}+\frac{a_{2}}{2^{3}}+\ldots$

$\frac{S}{2}=\frac{a_{1}}{2}+d\left(\frac{1}{2^{2}}+\frac{1}{2^{3}}+\ldots\right)$

$\frac{S}{2}=\frac{a_{1}}{2}+d\left(\frac{\frac{1}{4}}{1-\frac{1}{2}}\right)$

$\therefore S=a_{1}+d=a_{2}=4$

Or $4 a_{2}=16$

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