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 term is $\frac{1}{2}(mn+1)$

Answer 1

$T_m=\frac{1}{n}$ & $T_n=\frac{1}{m}$

$\Rightarrow a+(m-1)d=\frac{1}{n}$ & $a+(n-1)d=\frac{1}{m}$

$\Rightarrow na+mnd-nd=1$ & $ma+mnd-md=1$

equating the above $2$ equations, we get

$na+mnd-nd=ma+mnd-md$

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

$\Rightarrow a=d$

Substituting the value of $a=d$ in $na+mnd-nd=1$

We get , $\Rightarrow nd+mnd-nd=1$

$\Rightarrow d=\frac{1}{mn}=a$

$S_{mn}=\frac{mn}{2}[2a+(mn-1\left)d\right]$

$=\frac{mn}{2}[\frac{2}{mn}+\frac{(mn-1)}{mn}]$

$=\frac{mn}{2}\times \frac{1}{mn}[2+mn-1]$

$S_n\Rightarrow \frac{1}{2}(mn+1)$

Hence proved.