If the $$m^{th}$$ term of an arithmetic progression is $$\frac{1}{n}$$ and the $$n^{th}$$ term is $$\frac{1}{m}$$, then the $$mn^{th}$$ term of this progression will be

The $$m^{th}$$ term of an AP : [a + (m-1)(d)] = $$\dfrac{1}{n}$$

The $$n^{th}$$ term of an AP : [a + (n-1)(d)] = $$\dfrac{1}{m}$$

Subtracting the two equations :$$(d)(m-n)=\dfrac{(m-n)\ }{mn}$$ 
                                                 $$d=\dfrac{\ 1}{mn}$$

Substituting "d" value in any one of the equation, we get : $$a=\dfrac{\ 1}{mn}$$

So, the $$mn^{th}$$ term of AP : [a + (mn-1)(d)] = $$\dfrac{\ 1}{mn}+\dfrac{\ \left(mn-1\right)}{mn}$$ = 1