Question

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

Answer 1

Let the AP be denoted as

$a_1,a_2,a_3,\dots ,a_n,\dots$

with a common difference d.

Sum of first m terms:

$S_m=\frac{m}{2}(2a_1+(m-1\left)d\right)$

Sum of tfirst n terms:

$S_n=\frac{n}{2}(2a_1+(n-1\left)d\right)$

Given that the 2 sums are equal. Hence,

$\frac{m}{2}(2a_1+(m-1\left)d\right)$ $=\frac{n}{2}(2a_1+(n-1\left)d\right)$

$2a_{1}m+(m^2-m)d=2a_{1}n+(n^2-n)d$

$\therefore 2a_1=d(1-m-n)$

Sum of first (m+n) terms is given by

$S_{m+n}=\frac{m+n}{2}[2a_1+(m+n-1\left)d\right]$

Substituting

$2a_1=d(1-m-n)$,

$S_{m+n}=\frac{m+n}{2}\left[d\right(1-m-n)+(m+n-1\left)d\right]$

$=0$

Hence, proved that the sum of the first $(m+n)$ terms is $0$.