Question

If the sum of first $m$ terms of an $A.P.$ is $n$ and the sum of its first $n$ terms is $m$, then find the sum of its first $(m+n)$ terms.

Answer 1

Sum of $p$ terms of an A.P. is given by $=\frac{p}{2}[2a+(p-1)d]$

Given that sum of first $m$ terms of an A.P. is $n$, i.e.,

$S_m=n$

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

$\Rightarrow 2am+m^{2}d-md=2n.....\left(1\right)$

Also given that sum of first $n$ terms of an A.P. is $m$, i.e.,

$S_n=m$

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

$\Rightarrow 2an+n^{2}d-nd=2m.....\left(2\right)$

Subtracting ${eq}^n\left(2\right)$ from $\left(1\right)$, we have

$(2am+m^{2}d-md)-(2an+n^{2}d-nd)=2n-2m$

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

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

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

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

Now for sum of first $(m+n)$ terms-

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

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

Hence the sum of first $(m+n)$ terms is $-(m+n)$.