For an AP, if $ m $ times the mth term equals $ \mathrm{n} $ times the $ n $ th term, prove that $ (m+n) $ th term of the AP is zero. $ (m ≠ n) $.

Given: 

$m$ times the $m^{th}$ term of an AP is equal to $n$ times its $n^{th}$ term.

To do: 

We have to prove that \( (m+n) \) th term of the AP is zero.

Solution:

$n^{th}$ term of the AP $=t_n=a+(n−1)d$

$m^{th}$ term of the AP $=t_m=a+(m−1)d$

$(m+n)^{th}$ term of the AP $=a+[(m+n)−1]d$

According to the question,

$\Rightarrow m \times t_m=n \times t_n$

$\Rightarrow m[a+(m−1)d]=n[a+(n−1)d]$

$\Rightarrow m[a+(m−1)d]−n[a+(n−1)d]=0$

$\Rightarrow a(m−n)+d[(m+n)(m−n)−(m−n)]=0$

$\Rightarrow (m−n)[a+d((m+n)−1)]=0$

$\Rightarrow a+[(m+n)−1]d=0$

Hence proved.