Question

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

• A. $9$
• B. $5$
• C. $1$
• D. $noneofthese$

Answer 1

Answer: (A)

$9$

$S=(3+\frac{3}{4}+\frac{3}{4^2}+....)+(\frac{d}{4}+\frac{2d}{4^2}+\frac{3d}{4^3}+..)$

$S=3(1+\frac{1}{4}+\frac{1}{4^2}+..)+\left(s\right)$ (lets's say)

$s=\frac{d}{4}+\frac{2d}{4^2}+\frac{3d}{4^3}+..\left(1\right)$

$\frac{s}{4}=\frac{d}{4^2}+\frac{2d}{4^3}+...\left(2\right)$

subtrating (1)-(2)

$\frac{3s}{4}=\frac{d}{4}+\frac{d}{4^2}+\frac{d}{4^3}+....$

$s=\frac{d}{3}(1+\frac{1}{4}+\frac{1}{4^2}+..)$

$s=\frac{4d}{9}$

$S=3\left(\frac{4}{3}\right)+\frac{4d}{9}$

$S=4+\frac{4d}{9}=8$

$d=9$