Question

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

Answer 1

Let the first term be $a$ and common difference of the given
AP is $d.$
Given: $S_m=S_n$
$\Rightarrow \frac{m}{2}[2a+(m-1\left)d\right]=\frac{n}{2}[2a+(n-1\left)d\right]$
$\Rightarrow 2am+md(m-1)=2an+nd(n-1)$
$\Rightarrow 2am-2an+m^{2}d-md-n^{2}d+nd=0$
$\Rightarrow 2a(m-n)+d\left[\right(m^2-n^2)-(m-n\left)\right]=0$
$\Rightarrow 2a(m-n)+d\left[\right(m-n\left)\right(m+n)-(m-n\left)\right]=0$
$\Rightarrow (m-n)[2a+\{(m+n)-1\left\}d\right]=0$
$\Rightarrow 2a+(m+n-1)d=0[\because m-n\ne 0]...\left(i\right)$
Now,
$S_{m+n}=\frac{m+n}{2}[2a+(m+n)-1d]=0$
$\Rightarrow S_{m+n}=\frac{m+n}{2}\times 0$ [using $\left(i\right)$]
$\Rightarrow S_{m+n}=0$
Hence proved