Question

Suppose the sum of the first $m$ terms of an arithemetic progression is $n$ and the sum of its first $n$ terms is $m$, where $m\ne n$. Then the sum of the first $(m+n)$ terms of the arithemetic progression is

• A. $1-mn$
• B. $mn-5$
• C. $-(m+n)$
• D. $m+n$

Answer 1

The correct option is C $-(m+n)$

Let the first term and common difference of the A.P. be $a$ and $d$ respectively.
sum of first $m$ terms,
$S_m=\frac{m}{2}[2a+(m-1\left)d\right]=n$.......(i)
sum of first $n$ terms,
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]=m$.........(ii)

The sum of first $m+n$ terms,
$S_{m+n}=\frac{m+n}{2}[2a+(m+n-1\left)d\right]$

Finding ,
$S_m-S_n=(n-m)$
$\Rightarrow a(m-n)+\frac{d}{2}\left[m\right(m-1)-n(n-1\left)\right]=(n-m)$
$\Rightarrow a(m-n)+\frac{d}{2}[m^2-m-n^2+n]=(n-m)$
$\Rightarrow a(m-n)+\frac{d}{2}[m^2-n^2-m+n]=(n-m)$
$\Rightarrow a(m-n)+\frac{d}{2}\left[\right(m-n\left)\right(m+n-1\left)\right]=(n-m)$
$\therefore a+\frac{d}{2}(m+n-1)=-1$......(iii)

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