Question

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

Answer 1

Correct Answer: 9

To solve \(8 = 3 + \frac{1}{4}(3 + p) + \frac{1}{4^2}(3 + 2p) + \frac{1}{4^3}(3 + 3p) + \ldots\), we recognize a geometric series involving the common ratio \( \frac{1}{4} \). Let the series be \( S \), where \( S = 3 + \sum_{n=1}^{\infty} \frac{1}{4^n}(3+np) \). We split it into two separate series: \[ S = 3 + \sum_{n=1}^{\infty} \frac{3}{4^n} + \sum_{n=1}^{\infty} \frac{np}{4^n} \] The first series is geometric with first term \( a = \frac{3}{4} \) and common ratio \( r = \frac{1}{4} \). The sum of this infinite geometric series is: \[ \sum_{n=1}^{\infty} \frac{3}{4^n} = \frac{\frac{3}{4}}{1 - \frac{1}{4}} = \frac{3}{4} \times \frac{4}{3} = 1 \] The second series is an arithmetico-geometric series. Its sum can be found using the formula: \[ \sum_{n=1}^{\infty} \frac{n}{r^n} = \frac{r}{(1-r)^2} \text{, for } x = \frac{1}{4} \] Plugging in, we get: \[ \sum_{n=1}^{\infty} \frac{np}{4^n} = \frac{\frac{1}{4}}{(1 - \frac{1}{4})^2} \times p = \frac{\frac{1}{4}}{\frac{9}{16}} \times p = \frac{4}{9}p \] Substituting back into the equation, the series sum becomes: \[ S = 3 + 1 + \frac{4}{9}p = 8 \] Simplifying gives us: \[ 4 + \frac{4}{9}p = 8 \] \[ \frac{4}{9}p = 4 \] \[ p = 9 \] Thus, the value of \( p \) is 9. This value falls within the given range \([9,9]\), confirming our solution is correct. 

Answer 2

Given series:
\(8=3+\dfrac{1}{4}(3+p)+\dfrac{1}{4^2}(3+2p)+\dfrac{1}{4^3}(3+3p)+\ldots\)
This is an arithmetic-geometric progression (A.G.P). Using the sum formula for an infinite
A. G. P. ,  we have:
\(\text{Sum}=\frac{a}{1-r}+\frac{d\cdot r}{(1-r)^2}\)
Solving for \(p:\)
\(\dfrac{4p}{9}=4\Rightarrow p=9\)