Let $a$ and $d$ be the first term and the common difference of the $A.P.$ respectively.

Here,

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

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

According to the given condition, $\frac{p}{2}[2 a+(p-1) d]=\frac{q}{2}[2 a+(q-1) d]$

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

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

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

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

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

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

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

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

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

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

$=\frac{p+q}{2}[2 a-2 a]$

$=0$

Thus, the sum of the first $(p+q)$ terms of the $A.P.$ is $0$