If 8 = 3 + 14(3 + p) + 14^2(3 + 2p) + 14^3(3 + 3p) +, then the value of p is:

Given: 8 = 3 + 14(3 + p) + 14^2(3 + 2p) + 14^3(3 + 3p) + Find: p Let the series be S = 3 + n=1^ 14^n(3 + np) Then split it into two series: S = 3 + n=1^ 34^n + n=1^ np4^n For the geometric part, n=1^ 34^n = 341 - 14 = 1 For the arithmetico-geometric part, using n=1^ nx^n = x(1-x)^2 with x = 14, n=1^ np4^n = p 14 (1- 14)^2 = p 14 916 = 4p9 Substituting back, S = 3 + 1 + 4p9 = 8 So, 4 + 4p9 = 8 4p9 = 4 p = 9 Therefore, the value of p is 9. The correct option is A. Given: 8=3+ 14(3+p)+ 14^2(3+2p)+ 14^3(3+3p)+ Find: p This is an arithmetic-geometric progression. Using the infinite A.G.P. sum formula, Sum = a1-r + d r(1-r)^2 Applying it to the variable part gives directly 4p9 = 4 Hence, p = 9 Therefore, the value of p is 9. The correct option is A.

If 8 = 3 + 14(3 + p) + 14^2(3 + 2p) + 14^3(3 + 3p) +, then the va… — Solution | JEE Mathematics

MCQMediumJEE 2024Sum of Series

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

A
$9$
B
$8$
C
$7$
D
$6$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

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

Find: $p$

Let the series be

S = 3 + \sum_{n=1}^{\infty} \frac{1}{4^n}(3 + np)

Then split it into two series:

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

For the geometric part,

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

For the arithmetico-geometric part, using

\sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x)^2}

with $x = \frac{1}{4}$ ,

\sum_{n=1}^{\infty} \frac{np}{4^n} = p \cdot \frac{\frac{1}{4}}{\left(1-\frac{1}{4}\right)^2} = p \cdot \frac{\frac{1}{4}}{\frac{9}{16}} = \frac{4p}{9}

Substituting back,

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

So,

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

\frac{4p}{9} = 4

p = 9

Therefore, the value of $p$ is $9$ . The correct option is A.

Using A.G.P. Sum Formula

Given:

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

Find: $p$

This is an arithmetic-geometric progression. Using the infinite A.G.P. sum formula,

\text{Sum} = \frac{a}{1-r} + \frac{d \cdot r}{(1-r)^2}

Applying it to the variable part gives directly

\frac{4p}{9} = 4

Hence,

p = 9

Therefore, the value of $p$ is $9$ . The correct option is A.

Common mistakes

Treating the whole series as a simple geometric progression is incorrect because the bracketed term $3+np$ changes with $n$ . Split it into a geometric part and an arithmetico-geometric part instead.
Using the formula for $\sum x^n$ in place of $\sum nx^n$ is wrong for the second part. The correct identity is $\sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x)^2}$ for $|x|<1$ .
Starting the summation from the wrong index can change the constant term. Here, the standalone $3$ must be kept separate, and the summation begins from $n=1$ .

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