Question

The sum of $m$ terms is equal to $n$ and the sum of $n$ terms is equal to $m$, then prove that the sum of $(m+n)$ terms is $-(m+n)$.

Answer 1

$S_m=n,S_n=m$(given)

We know that the sum of ${}^{\prime }{n}^{\prime }$ terms is given by $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

Now, $\frac{m}{2}(2a+(m–1\left)d\right)=n$

$\Rightarrow 2am+m(m-1)d=2n\dots \dots \left(i\right)$

Similarily, we have

$\Rightarrow 2an+n(n-1)d=2m\dots \dots \left(ii\right)$

On subtracting eq.(ii) from (i), we get

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

$\Rightarrow 2am+m^{2}d-md-2an-n^{2}d+nd=2n-2m$

$\Rightarrow 2a(m-n)+m^{2}d-n^{2}d-md+nd=-2(m-n)$

$\Rightarrow 2a(m-n)+(m-n)(m+n)d-(m-n)d=-2(m-n)$

$\Rightarrow 2a+(m+n–1)d=–2$

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

$S_{m+n}=–(m+n)$