JEE MAIN - Mathematics (2024 - 27th January Morning Shift - No. 21)

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

Answer

$$8=\frac{3}{1-\frac{1}{4}}+\frac{p \cdot \frac{1}{4}}{\left(1-\frac{1}{4}\right)^2}$$

$$\text { (sum of infinite terms of A.G.P }=\frac{a}{1-r}+\frac{d r}{(1-r)^2} \text { ) }$$

$$\Rightarrow \frac{4 p}{9}=4 \Rightarrow p=9$$