Question

If the $m^{th}$ term of an A.P. be $\frac{1}{n}$ and $n^{th}$ term be $\frac{1}{m}$, then show that its $(mn{)}^{th}$ term is $1$.

Answer 1

Let $a$ and $d$ be the first term and common difference respectively of the given A.P.

Then,
$\Rightarrow \frac{1}{n}=m^{th}$ term $⟹\frac{1}{n}=a+(m-1)d$ ...(i)
$\Rightarrow \frac{1}{m}=n^{th}$ term $⟹\frac{1}{m}=a+(n-1)d$ ...(ii)
On subtracting equation (ii) from equation (i), we get
$\Rightarrow \frac{1}{n}-\frac{1}{m}=(m-n)d⟹\frac{m-n}{mn}=(m-n)d⟹d=\frac{1}{mn}$
Putting $d=\frac{1}{mn}$ in equation (i), we get
$\Rightarrow \frac{1}{n}=a+\frac{(m-1)}{mn}⟹\frac{1}{n}=a+\frac{1}{n}-\frac{1}{mn}⟹a=\frac{1}{mn}$
$\therefore (nm{)}^{th}$ term $=a+(nm-1)d=\frac{1}{mn}+(mn-1)\frac{1}{mn}=1$ $[\because a=\frac{1}{mn}=d]$