Let the first term be a and common difference be d. According to question $S_p = S_q$ ⇒ $\frac{p}{2}$ [2a + (p - 1) d] = $\frac{q}{2}$ [2a + (q - 1) d] ⇒ 2ap + $(p^2-p)d$ = 2aq + $(q^2-q)d$ ⇒ 2ap - 2aq + $(p^2-p)d$ - $(q^2-q)d$ = 0 ⇒ 2a (p – q) + d [$(p^2 - q^2)$ – (p – q)] = 0 ⇒ (p – q) [2a + d {(p + q) – 1}] = 0 ⇒ 2a + (p + q – 1) d = 0 (Since p – q ≠ 0) ....(i) ∴ $S_{p+q}$ = $\frac{p+q}{2}$ {2a + (p + q - 1) d} = $\frac{p+q}{2}$ (0) = 0 [from (i)]