Answer: Proved, \(S_{p+q} = 0\)
\[\frac{p}{2}[2a + (p - 1)d] = \frac{q}{2}[2a + (q - 1)d]\]\[2a(p - q) + d[(p^2 - q^2) - (p - q)] = 0\]\[(p - q)[2a + (p + q - 1)d] = 0\]Since \(p \neq q\), \(2a + (p + q - 1)d = 0\).
\[S_{p+q} = \frac{p + q}{2}[2a + (p + q - 1)d]\]\[= \frac{p + q}{2} \times 0 = 0\]