MCQ
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
- A$15$
- ✓$16$
- C$14$
- D$13$
$\frac{S}{2} =\frac{a_{1}}{2^{2}}+\frac{a_{2}}{2^{3}}+\ldots$
$\frac{S}{2}=\frac{a_{1}}{2}+d\left(\frac{1}{2^{2}}+\frac{1}{2^{3}}+\ldots\right)$
$\frac{S}{2}=\frac{a_{1}}{2}+d\left(\frac{\frac{1}{4}}{1-\frac{1}{2}}\right)$
$\therefore S=a_{1}+d=a_{2}=4$
Or $4 a_{2}=16$
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