If the sum of the first $p$ terms of an AP be $q$ and the sum of its first $q$ terms be $p$ then show that the sum of its first $(p + q)$ terms is $- (p + q)$.

1. Let $a$ be the first term and $d$ be the common difference of the given AP. Then, \begin{aligned} & S_ { p } = q \\ \implies & \dfrac { p } { 2 } (2a + (p - 1)d) = q \\ \implies & 2ap + p(p - 1)d = 2q && \ldots \text{(i)} \\ \end{aligned} And, \begin{aligned} & S_ { q } = p \\ \implies & \dfrac { q } { 2 } (2a + (q - 1)d) = p \\ \implies & 2aq + q(q - 1)d = 2p && \ldots \text{(ii)} \\ \end{aligned}
2. On subtracting $\text{(ii)}$ from $\text{(i)},$ we get \begin{aligned} & [ 2ap + p(p - 1)d ] - [ 2aq + q(q - 1)d ] = 2q - 2p \\ \implies & 2a(p - q) + (p^2 - p - q^2 + q)d = 2q - 2p \\ \implies & 2a(p - q) + (p^2 - q^2)d - (p - q)d = -2(p - q) \\ \implies & 2a(p - q) + (p - q)(p + q)d - (p - q)d = -2(p - q) \\ \implies & 2a + (p + q)d - d = -2 \\ \implies & 2a + (p + q - 1)d = -2 && \ldots \text{(iii)}\\ \end{aligned}
3. Now, the sum of the first $(p + q)$ terms of the AP is \begin{aligned} S_{ p + q } & = \dfrac{ p + q } { 2 } (2a + (p + q -1)d) \\ & = \dfrac{ p + q } { 2 } (-2) && \text{[Using (iii)]} \\ & = - (p + q) \\ \end{aligned}
4. Thus, the sum of the first $(p + q)$ terms is $- (p + q).$