# If in an AP , $\;m^{th}$ term is $\;\frac{1}{n}$ and $\;n^{th}\;term\;is\frac{1}{m}$ , then $\;mn^{th}$ term is :
$(a)\;\frac{1}{mn}\qquad(b)\;\frac{1}{m+n}\qquad(c)\;\frac{1}{m}+\frac{1}{n}\qquad(d)\;1$