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If the sum of first $m$ terms of an A.P. is same as sum of its first $n$ terms ($m \ne n$), then show that the sum of its first $(m + n)$ terms is zero.
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$S_m = S_n$
$\Rightarrow \frac{m}{2}[2a + (m - 1)d] = \frac{n}{2}[2a + (n - 1)d]$
$\Rightarrow 2a(m - n) = d(n^2 - m^2) - d(n - m)$
$\Rightarrow 2a = -d(m + n - 1)$
or $2a + (m + n - 1)d = 0$
i.e., $S_{m+n} = \frac{m+n}{2}[2a + (m + n - 1)d] = 0$
$\Rightarrow \frac{m}{2}[2a + (m - 1)d] = \frac{n}{2}[2a + (n - 1)d]$
$\Rightarrow 2a(m - n) = d(n^2 - m^2) - d(n - m)$
$\Rightarrow 2a = -d(m + n - 1)$
or $2a + (m + n - 1)d = 0$
i.e., $S_{m+n} = \frac{m+n}{2}[2a + (m + n - 1)d] = 0$