Question

If the $m^{th}$ and the $n^{th}$ term of an A.P are $\frac{1}{n}$ and $\frac{1}{m}$ respectively, then find its $m{n}^{th}$ term.

Answer 1

Given that, $m^{th}term=\frac{1}{n}$ and $n^{th}term=\frac{1}{m}.$
then, Let ${}^{\prime }{a}^{\prime }$ and ${}^{\prime }{d}^{\prime }$ be the first term and the common difference of the A.P.

We know that any term of an AP is given by $T_n=a+(n-1)d$
so, $a+(m-1)d=\frac{1}{n}\dots \dots \left(i\right)$ [$\because m^{th}term=\frac{1}{n}$, and $n^{th}term=\frac{1}{m}$]

and, $a+(n-1)d=\frac{1}{m}\dots \dots \left(ii\right)$
On subtracting equation $\left(ii\right)$ from $\left(i\right),$ we get

$a+(m-1)d-[a+(n-1\left)d\right]=\frac{1}{n}-\frac{1}{m}$
$\Rightarrow md-d-nd+d=\frac{1}{n}-\frac{1}{m}$
$\Rightarrow d(m-n)=\frac{m-n}{mn}$
$\Rightarrow d=\frac{1}{mn}$

Again if we put this value in equation (i), we get

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

$\Rightarrow a+m\times \frac{1}{mn}-\frac{1}{mn}=\frac{1}{n}$

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

Then, the $m{n}^{th}$ term of the AP is
$=a+(mn-1)d$

$=\frac{1}{mn}+(mn-1)\times \frac{1}{mn}$

$=\frac{1}{mn}+1-\frac{1}{mn}$

$=1$

Hence, $m{n}^{th}$ term of the given AP is $1.$