Question

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

Answer 1

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

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

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

Also, sum of first $n$ terms is $m$.

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

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

Subtracting $\left(2\right)$ from $\left(1\right)$,

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

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

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

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

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