If the sum of first $m$ terms of an A.P. is same as sum of its first $n$ terms $(m
eq n)$, then show that the sum of its first $(m + n)$ terms is zero.

Question

If the sum of first $m$ terms of an A.P. is same as sum of its first $n$ terms $(m 
eq n)$, then show that the sum of its first $(m + n)$ terms is zero.

Accepted Answer

Step 1: $S_m = S_n \Rightarrow \frac{m}{2}[2a + (m - 1)d] = \frac{n}{2}[2a + (n - 1)d]$
Step 2: $\Rightarrow 2a(m - n) = d(n^2 - m^2) - d(n - m)$
Step 3: $\Rightarrow 2a = -d(m + n - 1)$, i.e., $2a + (m + n - 1)d = 0$
Step 4: $S_{m+n} = \frac{m + n}{2}[2a + (m + n - 1)d] = \frac{m + n}{2} 	imes 0 = 0$

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