If 8=3+\frac14(3+p)+\frac14^2(3+2 p)+\frac14^3(3+3 p)+\cdots \cdots \infty , then the value of p is \underline\hspace2cm .

Sequences and Series PYQ — JEE Main 2024 · 27 Jan · Shift 1 · Q51 with Solution

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Correct answer: 9

Given

8=3+\frac{1}{4}(3+p)+\frac{1}{4^2}(3+2p)+\frac{1}{4^3}(3+3p)+\cdots

Observe that after the first term, the general term is

\frac{1}{4^n}(3+np),\qquad n\ge 1

So,

8=3+\sum_{n=1}^{\infty}\frac{3+np}{4^n}

8=3+3\sum_{n=1}^{\infty}\frac{1}{4^n}+p\sum_{n=1}^{\infty}\frac{n}{4^n}

For $|r|<1$ ,

\sum_{n=1}^{\infty} r^n=\frac{r}{1-r}, \qquad \sum_{n=1}^{\infty} nr^n=\frac{r}{(1-r)^2}

Here $r=\frac14$ .

So,

\sum_{n=1}^{\infty}\frac{1}{4^n}=\frac{\frac14}{1-\frac14}=\frac{1}{3}

and

\sum_{n=1}^{\infty}\frac{n}{4^n}=\frac{\frac14}{\left(1-\frac14\right)^2} =\frac{\frac14}{\left(\frac34\right)^2} =\frac{\frac14}{\frac{9}{16}} =\frac{4}{9}

8=3+3\left(\frac13\right)+p\left(\frac49\right)

8=3+1+\frac{4p}{9}

8=4+\frac{4p}{9}

4=\frac{4p}{9}

p=9

The required integer is

\boxed{9}

Stored correct answer = $9$ .

My derived answer also is $9$ , so they agree.