Question

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

Answer 1

We know,

$a_n=a+(n-1)d$

According to the question,

$m\left(a_m\right)=n\left(a_n\right)$

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

$\Rightarrow am+(m-1)md=an+(n-1)d$

$\Rightarrow am+m^{2}d-md=an+n^{2}d-nd$

$\Rightarrow am-an=n^{2}d-nd-m^{2}d+md$

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

$\Rightarrow a(m-a)=d(m+n)(n-m)+d(m-n)$

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

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

$\Rightarrow a=d(1-m-n)(\because m\ne n)$

$\Rightarrow a=d(1-m-n)...\left(1\right)$

Now,

$a_{m+n}=(a+(m+n-1\left)d\right)$

$=\left(\right(1-m-n)d+(m+n-1\left)d\right)\left(from\right(1\left)\right)$

$=d(1-m-n+m+n-1)$

$=0$

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