If the sum of p terms of an A.P. is q and the sum of q terms is p, then the sum of p + q terms will be
(a) 0
(b) p − q
(c) p + q
(d) − (p + q)
(d) $-(p+q)$
$S_{p}=q$
$\Rightarrow \frac{p}{2}\{2 a+(p-1) d\}=q$
$\Rightarrow 2 a p+(p-1) p d=2 q$ .....(1)
$S_{q}=p$
$\Rightarrow \frac{q}{2}\{2 a+(q-1) d\}=p$
$\Rightarrow 2 a q+(q-1) q d=2 p$ ....(2)
Multiplying equation (1) by $q$ and equation (2) by $p$ and then solving, we get:
$d=\frac{-2(p+q)}{p q}$
Now, $S_{p+q}=\frac{(p+q)}{2}[2 a+(p+q-1) d]$
$=\frac{p}{2}[2 a+(p-1) d+q d]+\frac{q}{2}[2 a+(q-1) d+p d]$
$=S_{p}+\frac{p q d}{2}+S_{q}+\frac{p q d}{2}$
$=p+q+p q d$
$=p+q-\frac{2(p+q) p q}{p q}$
$=-(p+q)$
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