• # question_answer If the sum of first $n$ terms of an A.P. be equal to the sum of its first $m$ terms, $(m\ne n)$, then the sum of its first $(m+n)$ terms will be [MP PET 1984] A)   0 B) $n$ C) $m$ D) $m+n$

Solution :

As given  $\frac{n}{2}\left\{ 2a+(n-1)d \right\}=\frac{m}{2}\left\{ a+(m-1)d \right\}$ $\Rightarrow$$2a(m-n)+d({{m}^{2}}-m-{{n}^{2}}+n)=0$ $\Rightarrow$ $(m-n)\left\{ 2a+d(m+n-1) \right\}=0$ $\Rightarrow$ $2a+(m+n-1)d=0$, $(\because \ m\ne n)$ $\therefore$ ${{S}_{m+n}}=\frac{m+n}{2}\left\{ 2a+(m+n-1)d \right\}=\frac{m+n}{2}\left\{ 0 \right\}=0$.

You need to login to perform this action.
You will be redirected in 3 sec