If the sum of p terms of an A.P. is q and the sum of q terms is p,

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

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

$=S_{p}+\frac{p q d}{2}+S_{q}+\frac{p q d}{2}$

$=p+q+p q d$

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

$=-(p+q)$