Question

If $3+\frac{1}{4}(3+P)+\frac{1}{4^2}(3+2P)+\frac{1}{4^3}(3+3P)+\dots =8,$ then the value of $P$ is

• A. $9$
• B. $15$
• C. $21$
• D. $12$

Answer 1

The correct option is A $9$

Given series is
$3+\frac{1}{4}(3+P)+\frac{1}{4^2}(3+2P)+\frac{1}{4^3}(3+3P)+\dots$
It can be written as
$1\cdot 3+\frac{1}{4}(3+P)+\frac{1}{4^2}(3+2P)+\frac{1}{4^3}(3+3P)+\dots$
So, the given series is AGP with $a=3,r=\frac{1}{4},d=p$
For $|r|<1,n\to \infty$
$S_{\infty }=\frac{a}{1-r}+\frac{dr}{(1-r{)}^2}$
$\Rightarrow \frac{3}{1-\frac{1}{4}}+\frac{p\cdot \frac{1}{4}}{{(1-\frac{1}{4})}^2}=8$
$\Rightarrow 4+\frac{4p}{9}=8$
$\Rightarrow p=9$

Alternate Solution:
$3+\frac{1}{4}(3+P)+\frac{1}{4^2}(3+2P)+\frac{1}{4^3}(3+3P)+\dots =8$
$\Rightarrow 3(1+\frac{1}{4}+\frac{1}{4^2}+\dots )+P(\frac{1}{4}+\frac{2}{4^2}+\dots )=8\dots \left(1\right)$

Now Let $x=\frac{1}{4}+\frac{2}{4^2}+\dots$
$\Rightarrow \frac{x}{4}=\frac{1}{4^2}+\frac{2}{4^3}+\dots$
Then $x-\frac{x}{4}=\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\dots$
$\Rightarrow x=\frac{4}{3}(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\dots )$

From equation $\left(1\right)$
$3(1+\frac{1}{4}+\frac{1}{4^2}+\dots )+P\cdot \frac{4}{3}(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\dots )=8$
$\Rightarrow 3\cdot \frac{1}{1-\frac{1}{4}}+P\cdot \frac{4}{3}\cdot \frac{\frac{1}{4}}{1-\frac{1}{4}}=8$
$\Rightarrow 4+\frac{4P}{9}=8$
$\Rightarrow P=9$