Question

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

Answer 1

Let the first term of AP =a
common differnce =d.
Sum of first n terms =$\frac{n}{2}(2a+(n-1\left)d\right)=m$ given.
$\therefore 2an+n(n-1)d=2m.-\left(1\right)$
Similary Sum of first m term =$\frac{m}{2}[2a+(m-1\left)d\right]=n$
2am+m(m-1)d =2n. -(2)
Subtract (2) from (1)
2a(n-m) +n(n-1)d-m(m-1)d=2(m-n)
$2a(n-n)+d\{n^2-n-m^2+n\}=2(m-n)$
2a(n-m)+d{(n+m)(n-m)-(n-m}=2(m-n)
Take out n -m common,
2a+d{n+m-1}=-2 -(3)
Sum of (m+n) terms =$\frac{m+n}{2}[2a+(m+n-1\left)d\right]$
$=\frac{m+1}{2}[-2]$ (from 3)
$S_{m+n}=-(m+n)$-proved.