Question

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

Answer 1

$\Rightarrow$Let be first term $=a$

$\Rightarrow$common difference $=d$

$\Rightarrow$$T_m=a+(m-1)d....\left(i\right)$

$\Rightarrow$$T_n=a+(n-1)d....\left(ii\right)$

According to question,

$\Rightarrow$$m{T}_m=n{T}_n$

$\Rightarrow m[a+(m-1\left)d\right]=n[a+(n-1\left)d\right]$

$\Rightarrow ma+m(m-1)d=na+n(n-1)d$

$\Rightarrow ma-na+m(m-1)d-n(n-1)d=0$

$\Rightarrow a(m-n)+d[m^2-m-n^2+n]=0$

Put the value of $a$ in equation $\left(iii\right)$

$\Rightarrow$$T_{m+n}=-d(m+n-1)+(m+n-1)d$

$\Rightarrow$$T_{m+n}=0$

$\therefore (m+n)th$ term of A.P. is zero.