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 ${\left(mn\right)}^{th}$ term is 1.

Answer 1

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

$\frac{1}{n}=m^{th}$ term

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

$\frac{1}{m}=n^{th}$ term

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

On subtracting equation (ii) from equation (i), we get

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

$\frac{m-n}{mn}=(m-n)d$

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

Putting $d=\frac{1}{mn}$ in equation (i) , we get

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

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

${mn}^{th}$ term $=a+(mn-1)d$

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