Question

Suppose the sum of the first $m$ terms of an arithmetic 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 arithmetic progression is

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

Answer 1

Answer: (C)

Given, $S_m=n$ and $S_n=m$
$S_m=\frac{m}{2}[2a+(m-1)d]=n...(i)$
$S_n=\frac{n}{2}(2a+(n-1)d)=m...(ii)$
On subtracting Eq. $(ii)$ from Eq. $(i)$, we get
$(m-n)a+(m-n)(m+n-1)\frac{d}{2}$
$=-(m-n)$
$\Rightarrow 2a+(m+n-1)d=-2[m\ne n]$
$\therefore s_{m+n}=\frac{m+n}{2}(2a+(m+n-1)d)$
$=\frac{m+n}{2}(-2)=-(m+n)$