Question

If $m^{th}$ term of an AP is $\frac{1}{n}$ and $n^{th}$ term is $\frac{1}{m}$ then find the sum of its first $mn$ terms.

Answer 1

We know that the sum of the $n^{th}$ of an A.P is given by $a_n=a+(n-1)d$

Given: $a_m=\frac{1}{n}$

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

$\Rightarrow an+mnd-nd=1\dots \dots \left(i\right)$

and, it is given that, $a_n=\frac{1}{m}$

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

$\Rightarrow am+mnd-md=1\dots \dots \left(ii\right)$

From $eq.\left(i\right)$ and $\left(ii\right),$ we get

$an+mnd-nd=am+mnd-md$

$\Rightarrow an-am-nd+md=0$

$\Rightarrow a(n-m)-(n-m)d=0$

$\Rightarrow a(n-m)=(n-m)d$

$\therefore a=d$

On putting $a=d$ in $eq.\left(i\right)$, we get

$nd+mnd-nd=1$ [on substituting $a=d$]

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

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

Sum of $mn$ terms of the given A.P is $S_{mn}=\frac{mn}{2}[2a+(mn-1\left)d\right]$

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

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

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