In an arithmetic sequence, the sum of the first $m$ terms is equal to the sum of the first $n$ terms.
Prove that the sum of the first $m + n$ terms is equal to zero.
Let first term of the arithmetic sequence $ = a$
and common difference $= d$
Sum of first $m$ terms of the arithmetic sequence $= S_m = \dfrac{m}{2} \left[2a + \left(m - 1\right) d \right]$
Sum of first $n$ terms of the arithmetic sequence $= S_n = \dfrac{n}{2} \left[2a + \left(n - 1\right) d \right]$
Given: $\;$ $S_m = S_n$
i.e. $\;\;\;$ $\dfrac{m}{2} \left[2a + \left(m - 1\right) d \right] = \dfrac{n}{2} \left[2a + \left(n - 1\right) d \right]$
i.e. $\;\;\;$ $2am + m \left(m - 1\right) d = 2an + n \left(n - 1\right) d$
i.e. $\;\;\;$ $2a \left(m - n\right) = \left(n^2 - n - m^2 + m\right) d$
i.e. $\;\;\;$ $2a \left(m - n\right) = \left[\left(n^2 - m^2\right) - \left(n - m\right) \right] d$
i.e. $\;\;\;$ $2a \left(m - n\right) = \left[\left(n + m\right) \left(n - m\right) - \left(n - m\right) \right] d$
i.e. $\;\;\;$ $\left(n - m\right) \left(n + m - 1\right) d - 2a \left(m - n\right) = 0$
i.e. $\;\;\;$ $\left(n - m\right) \left[2a + \left(m + n - 1\right)d\right] = 0$
$\because$ $\;\;\;$ $n \neq m \implies n - m \neq 0$
$\therefore$ $\;\;\;$ $2a + \left(m + n - 1\right) d = 0$ $\;\;\; \cdots \; (1)$
Now, $\;$ $S_{m + n} = \left(\dfrac{m + n}{2}\right) \left[2a + \left(m + n - 1\right) d \right]$ $\;\;\; \cdots \; (2)$
$\therefore$ $\;\;\;$ In view of equation $(1)$, equation $(2)$ becomes
$S_{m + n} = \left(\dfrac{m + n}{2}\right) \times 0$
i.e. $\;\;\;$ $S_{m + n} = 0$
Hence proved.