If the sum of an A.P. is the same for $p$ as for $q$ terms, show that its sum for $p + q$ terms is zero.

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Solution

Sum of $p$ terms of an AP with first term $a$ and common difference $d$ is
$S = \frac{p}{2} (2 a + (p - 1) d)$ ....... $(i)$

Sum of $q$ terms $= S = \frac{q}{2} (2 a + (q - 1) d)$ ....... $(i i)$

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

$\Rightarrow 2 a p + (p^{2} - p) d = 2 a q + (q^{2} - q) d$

$\Rightarrow 2 a (p - q) + d (p^{2} - q^{2}) - d (p - q) = 0$

$\Rightarrow (p - q) (2 a + d (p + q - 1)) = 0$

CASE 1:

$p - q = 0$

$\Rightarrow p = q$ But, $p \neq q$

Case invalid

CASE 2:

$2 a + d (p + q - 1) = 0$
Sum of $p + q$ terms $= \frac{p + q}{2} (2 a + (p + q - 1) d)$

$= \frac{p + q}{2} (0) = 0$

Hence, sum of $p + q$ terms $= 0$

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