Let $a$ be the first term and $d$ be the common difference of A.P.
Sum of first $p$ terms $=\frac{p}{2}[2 a+(p-1) d] \quad \ldots(i)$
Sum of first $q$ terms $=\frac{q}{2}[2 a+(q-1) d] \quad \ldots(ii)$
Equating (i) \& (ii)
$\frac{p}{2}[2 a+(p-1) d]=\frac{q}{2}[2 a+(q-1) d]$
Transposing the term of R.H.S to L.H.S
$\begin{array}{ll}\text { or } & 2 a(p-q)+p(p-1)-q(q-1) d=0 \\ \Rightarrow & 2 a(p-q)+\left[\left(p^2-q^2-(p-q) d\right]=0\right. \\ \text { or } & 2 a(p-q)+(p-q)[(p+q)-d]=0 \\ \Rightarrow & (p-q)[2 a+(p+q-1) d]=0 \\ \Rightarrow & 2 a+(p+q-1) d=0 \quad \ldots(iii)\end{array}$
$(\because p \neq q)$
Sum of first $(p+q)$ term
$\begin{aligned}
&=\frac{p+q}{2}[2 a+(p+q-1) d]=\frac{p+q}{2} \times 0=0 \\
&\therefore \quad 2 a+(p+q-1) d=0 \quad[\text { from (iii) }]
\end{aligned}$