Question

If the sum of first $m$ terms of an AP be $n$ and the sum of its first $n$ terms be $m$ then show that the sum of its first $(m+n)$ terms is $-(m+n).$

Answer 1

Let $a$ be the first term and $d$ be the common difference of the given AP. Then,

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

Given,

$S_m=n$

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

$2am+m(m-1)d=2n$ ...........(1)

$S_n=m$

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

$2an+n(n-1)d=2m$ ..........(2)

On subtracting 2 from 1, we get,

$2a(m-n)+\left[\right(m^2-n^2)-(m-n\left)\right]d=2(n-m)$

$(m-n)[2a+(m+n-1\left)d\right]=2(n-m)$

$2a+(m+n-1)d=-2$ ..........(3)

Sum of $(m+n)$ terms of the given AP

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

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

Hence proved.