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In an A.P., the first term is 1 and sum of the first p terms is 0, then sum of the first $(p + q)$ terms is

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$\frac{q (p + q)}{1 - p}$

$\frac{q}{1 - p}$

$\frac{(p + q) p}{1 - p}$

$\frac{2 (p + q)}{1 - p}$

answer is A.

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Detailed Solution

$0 = S_{p} = \frac{p}{2} [2 a + (p - 1) d] \Rightarrow d = \frac{2}{1 - p}$

$S_{p + q} = \frac{p + q}{2} [2 a + (p + q - 1) d]$

$= \frac{1}{2} (p + q) [2 a + (p - 1) d + q d]$

$= \frac{(p + q) q}{1 - p}$

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