Question

If $m$ times the $m^{th}$ of an A.P. is equal to $n$ times its $n^{th}$ term, then show that the $(m+n{)}^{th}$ term of the A.P. is zero

Answer 1

The $m$th term of the A.P is $T_m$ and the $n$th term is $T_n$.

It is given that $m$ times the $m$th term of an A.P is equal to $n$ times the $n$th term that is:

$m\left[T_m\right]=n\left[T_n\right]......\left(1\right)$

We know that the general term of an arithmetic progression with first term $a$ and common difference $d$ is $T_n=a+(n-1)d$, therefore, equation 1 can be rewritten as:

$m\left[T_m\right]=n\left[T_n\right]\Rightarrow m[a+(m-1\left)d\right]=n[a+(n-1\left)d\right]\Rightarrow ma+md(m-1)=na+nd(n-1)\Rightarrow ma+m^{2}d-md=na+n^{2}d-nd\Rightarrow ma-na+m^{2}d-n^{2}d-md+nd=0$

$\Rightarrow a(m-n)+(m^2-n^2-m+n)d=0\Rightarrow a(m-n)+\left[\right(m+n\left)\right(m-n)-(m-n\left)\right]d=0(\because (a+b\left)\right(a-b){=a}^2-b^2)\Rightarrow a(m-n)+(m-n)\left[\right(m+n)-1]d=0\Rightarrow a+\left[\right(m+n)-1]d=0\left(Divideby\right(m-n\left)\right)\Rightarrow T_{(m+n)}=0(\because T_n=a+(n-1\left)d\right)$

Hence, the $(m+n)$th term of the A.P is zero.