If 8 = 3 + 1 4 ( 3 + p ) + 1 4 2 ( 3 + 2 p ) + 1 4 3 ( 3 + 3 p ) + ? ? , then the value of p is _______ . [2024]

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Solution

Q.
If $8 = 3 + \frac{1}{4} (3 + p) + \frac{1}{4^{2}} (3 + 2 p) + \frac{1}{4^{3}} (3 + 3 p) + \dots \infty,$ then the value of $p$ is _______ . [2024]

Ans.
(9)

$8 = 3 + \frac{1}{(4)} (3 + p) + \frac{1}{{(4)}^{2}} (3 + 2 p) + \frac{1}{{(4)}^{3}} (3 + 3 p) + \dots \infty$ ...(i)

Multiplying both sides by $\frac{1}{4},$ we get $2 = \frac{3}{4} + \frac{3 + p}{{(4)}^{2}} + \frac{3 + 2 p}{{(4)}^{3}} + \dots + \infty$ ...(ii)

Subtracting (ii) from (i), we get $6 = 3 + \frac{p}{4} + \frac{p}{4^{2}} + \dots$

$\Rightarrow 3 = p [\frac{1}{4} + \frac{1}{{(4)}^{2}} + \frac{1}{{(4)}^{3}} + \dots + \infty]$

$\Rightarrow 3 = p [\frac{\frac{1}{4}}{1 - \frac{1}{4}}]$ $[∵ S_{\infty} = \frac{a}{1 - r}$ for infinite geometric series $]$

$\Rightarrow 3 = p [\frac{1}{4} \times \frac{4}{3}] \Rightarrow p = 9$

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