The $m^{th}$ term of an arithmetic progression is $x$ and the $n^{th}$ term is $y$. Then find the sum of the first $( m+n)$ terms.

Given: The $m^{th}$ term of an arithmetic progression is $x$ and the $n^{th}$  term is $y$.

To do: To find the sum of the first $( m+n)$ terms.

Solution: 

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

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

$x-y=[a+( m-1)d]-[a+( n-1)d]$

$x-y=( m-n)d$

$\therefore d=\frac{x-y}{m-n}$

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

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

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

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

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

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