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JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Sequences and Series

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 ____.

Answer: 9

Solution

The series is

k = 0 \sum \infty (\frac{1}{4})^{k} (3 + k p) = 3 \sum (\frac{1}{4})^{k} + p \sum k (\frac{1}{4})^{k}

.

= \frac{3}{1 - 1/4} + p \cdot \frac{1/4}{( 1 - 1/4 ) ^{2}} = 4 + \frac{4 p}{9}

Setting this equal to

8

:

\frac{4 p}{9} = 4 \Rightarrow p = 9

.

JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Integer answer

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