Question

If the $m^{th}$ term of an A.P. is $\frac{1}{n}$ and the $n^{th}$ term is $\frac{1}{m}$, show that the sum of mn terms is $\frac{1}{2}(mn+1)$

Answer 1

$T_m=a+(m-1)d=\frac{1}{n}{T}_n=a+(n-1)d=\frac{1}{m}$

$s_{mnn}=\frac{nm}{2}[a+a+(nm-1\left)d\right]s_{mnn}=\frac{nm}{2}[2\frac{1}{m}-(n-1)d+(nm-1\left)d\right]s_{mnn}=\frac{nm}{2}[2\frac{1-m(n-1)d}{m}+(nm-1\left)d\right]s_{mnn}=\frac{nm}{2}\left[\frac{2(1-mnd-md)+m^{2}nd-md}{m}\right]s_{mnn}=\frac{n}{2}[2-2mnd-2md+m^{2}nd-md]s_{mnn}=\frac{nm+1}{2}$