Question

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

Answer 1

We know,
$a_n=a+\left(n-1\right)d$

According to the question,

$$\begin{gathered} m\left(a_m\right)=n\left(a_n\right) \\ \Rightarrow m\left(a+\left(m-1\right)d\right)=n\left(a+\left(n-1\right)d\right) \\ \Rightarrow am+\left(m-1\right)md=an+\left(n-1\right)nd \\ \Rightarrow am+m^{2}d-md=an+n^{2}d-nd \\ \Rightarrow am-an=n^{2}d-nd-m^{2}d+md \\ \Rightarrow a\left(m-n\right)=d\left(n^2-m^2\right)+d\left(m-n\right) \\ \Rightarrow a\left(m-n\right)=d\left(m+n\right)\left(n-m\right)+d\left(m-n\right) \\ \Rightarrow a\left(m-n\right)=d\left[\left(m+n\right)\left(n-m\right)+\left(m-n\right)\right] \\ \Rightarrow a\left(m-n\right)=d\left[-\left(m+n\right)\left(m-n\right)+\left(m-n\right)\right] \\ \Rightarrow a\left(m-n\right)=d\left(m-n\right)\left[1-m-n\right] \\ \Rightarrow a=d\left(1-m-n\right)\left(\because m\ne n\right) \\ \Rightarrow a=d\left(1-m-n\right)...\left(1\right) \end{gathered}$$

Now,
$$\begin{gathered} a_{m+n}=\left(a+\left(m+n-1\right)d\right) \\ =\left(\left(1-m-n\right)d+\left(m+n-1\right)d\right)\left(\mathrm{from}\left(1\right)\right) \\ =d\left(1-m-n+m+n-1\right) \\ =0 \end{gathered}$$

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