Question

if $3+\frac{1}{4}(3+d)+\frac{1}{4^2}(3+2d)+$... to $\infty$ $=8$,then the value of $d$ is

• A. $9$
• B. $5$
• C. $1$
• D. none of these

Answer 1

The correct option is A $9$

Let $S=3+\frac{1}{4}(3+d)+\frac{1}{4^2}(3+2d)+...\to \left(1\right)$
Multiplying $\left(1\right)$ by $\frac{1}{4}$
We get
$\frac{1}{4}S=\frac{3}{4}+\frac{1}{4^2}(3+d)+\frac{1}{4^3}(3+3d)+...\to \left(2\right)$
$\left(1\right)-\left(2\right)$ gives
$(1-\frac{1}{4})S=3+\frac{d}{4}(1+\frac{1}{4}+\frac{1}{4^2}+...)$
$\Rightarrow \frac{3}{4}S=3+\frac{d}{4}\left(\frac{1}{1-\frac{1}{4}}\right)\Rightarrow \frac{3}{4}S=3+\frac{d}{4}\left(\frac{4}{3}\right)$
$\Rightarrow \frac{3}{4}S=3+\frac{d}{3}\Rightarrow \frac{3}{4}S=\frac{9+d}{3}\Rightarrow S=\left(\frac{9+d}{3}\right)\frac{4}{3}$
$\therefore 8=\left(\frac{9+d}{3}\right)\frac{4}{3}\Rightarrow d=9$