Question

If m times $m^{th}$ term of an $A.P.$ is equal to $n$ times its $n^{th}$ term find its $(m+n{)}^{th}$ term

Answer 1

Given,

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

$am+(m^2-m)d=an+(n^2-n)d$

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

$a(m-n)+d\left[\right(m+n\left)\right(m-n)-(m-n\left)\right]=0$

$(m-n)[a+d(m+n-1\left)\right]=0$

$m-n\ne 0$

$\Rightarrow a+d\left[\right(m+n)-1]=0$

Now,

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

$=0$