Question

If in an $A.P.$ the sum of $m$ terms is equal to $n$ and the sum of $n$ terms is equal to $m$, then prove that sum of $(m+n)$ terms is $-(m+n)$.

Answer 1

Here,

First term $=a$

Common difference $=d$

Given:

$S_m=n$

$\Rightarrow \frac{m}{2}(2a+(m-1)d)=n$ …… (1)

$S_n=m$

$\Rightarrow \frac{n}{2}(2a+(n-1)d)=m$ ……. (2)

Subtract equation (1) from (2).

$2a(n-m)+d(n^2-n-m^2+m)=2m-2n$

$2a(n-m)+d(n^2-n-+mn-mn-m^2+m)=-2(n-m)$

$(n-m)(2a+(n+m-1)d)=-2(n-m)$

$2a+(n+m-1)d=-2$

Now,

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

$S_{m+n}=\frac{(m+n)}{2}\times (-2)$

$S_{m+n}=-(m+n)$

Hence, proved.