In an A.P, $S_p=q$ , $S_q=p$   and $S_r$  denotes the sum of first r terms. Then, $S_{p+q}$  is equal to

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

$\Rightarrow 2ap+\left(p^2-p\right)d=2q$

$\Rightarrow 2ap+\left(p^2-p\right)d-2q=0\to \left(i\right)$

$S_q=p$

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

$\Rightarrow 2aq+\left(q^2-q\right)d-2p=0\to \left(ii\right)$

From (i) and (ii),

$2ap+\left(p^2-p\right)d-2q=2aq+\left(q^2-q\right)d-2p$

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

$\Rightarrow 2a+\left(p+q-1\right)d=-2\to \left(iii\right)$

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

$=\frac{p+q}{2}\times \left(-2\right)\left(from\left(iii\right)\right)$

$=-\left(p+q\right)$