Suppose m th term of an A.P. is 1/n and nth term of the A.P. is 1/m. If rth term of the A.P. is 1, then r is equal to

Suppose $m$ th term of an A.P. is 1/n and $n$ th term of the $A . P .$ is 1/m. If $r$ th term of the $A . P .$ is 1, then $r$ is equal to

Let $d$ be the common difference of the A.P., then

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

$\Rightarrow d = \frac{1}{m n}$

Now $a_{r} - a_{m} = (r - m) d$

$\begin{array}{l} \Rightarrow 1 = a_{r} = \frac{1}{n} + (r - m) \frac{1}{m n} = \frac{r}{m n} \\ \Rightarrow r = m n . \end{array}$

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