Question

If the $m$th term of an AP is $a$ and its $n$th term is $b$, show that the sum of its $(m+n)$ terms is
$\frac{(m+n)}{2}\{a+b+\frac{(a-b)}{(m-n)}\}$

Answer 1

Given,

$a+(m-1)d=a$

$a+(n-1)d=b$

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

$a-b=(m-n)d$

$\therefore d=\frac{a-b}{m-n}$

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

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

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

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

$=\frac{m+n}{2}[a+b+d]$

$\therefore S_n=\frac{m+n}{2}[a+b+\frac{a-b}{m-n}]$

Hence proved.