( \dfrac{1}{a},\ \dfrac{1}{b},\ \dfrac{1}{c} ) are in A.P.

[ 2\cdot\dfrac{1}{b} = \dfrac{1}{a} + \dfrac{1}{c} ] [ \dfrac{2}{b} = \dfrac{a+c}{ac} ] [ 2ac = b(a + c) ] [ b = \dfrac{2ac}{a + c} ]

Proved.