### If the sum of the first $p$ terms of an AP is the same as the sum of its first $q$ terms (where $p \ne q$) then show that the sum of its first $(p + q)$ terms is zero.

1. We know that the sum of first $n$ terms of an AP is given by $S_n = \dfrac { n } { 2 } (2a + (n-1)d),$ where $a$ is the first term and $n$ is the number of terms in the AP.
2. We are given that \begin{aligned} & S_p = S_q \\ \implies & \dfrac { p } { 2 } (2a + (p - 1)d) = \dfrac { q } { 2 } (2a + (q - 1)d) \\ \implies & p(2a + (p - 1)d) = q(2a + (q - 1)d) \\ \implies & 2ap + (p - 1)dp = 2aq + (q - 1)dq \\ \implies & 2ap - 2aq = (q - 1)dq - (p - 1)dp \\ \implies & 2a(p - q) = q^2d - dq - p^2d + dp \\ \implies & 2a(p - q) = q^2d - p^2d + dp - dq \\ \implies & 2a(p - q) = - d(p^2 - q^2) + d(p - q) \\ \implies & 2a(p - q) = - d(p - q)(p + q) + d(p - q) \\ \implies & 2a(p - q) = (p - q) [- d(p + q) + d] \\ \implies & 2a = - d(p + q) + d \\ \implies & 2a = (1 - p - q)d && \ldots \text{(i)} \\ \end{aligned}
3. Now, the sum of first $(p + q)$ terms of the given AP is \begin{aligned} S_{ p+q } & = \dfrac { p + q } { 2 } (2a + (p + q -1)d) \\ & = \dfrac { p + q } { 2 } ((1 - p - q)d + (p + q - 1)d) && \text{ [Using(i)] } \\ & = \dfrac { p + q } { 2 } (d - pd - qd + pd + qd - d) \\ & = \dfrac { p + q } { 2 } (0) \\ & = 0 \end{aligned}
4. Hence, the sum of $(p + q)$ terms is $0$ . 