Sequences And Series Question 6
Question 6 - 2024 (27 Jan Shift 1)
If
$8=3+\frac{1}{4}(3+p)+\frac{1}{4^{2}}(3+2 p)+\frac{1}{4^{3}}(3+3 p)+\ldots \infty$,
then the value of $p$ is
Show Answer
Answer (9)
Solution
$8=\frac{3}{1-\frac{1}{4}}+\frac{p \cdot \frac{1}{4}}{\left(1-\frac{1}{4}\right)^{2}}$
(sum of infinite terms of A.G.P $=\frac{a}{1-r}+\frac{d r}{(1-r)^{2}}$ )
$\Rightarrow \frac{4 p}{9}=4 \Rightarrow p=9$
