Given (\dfrac{1}{a},\ \dfrac{1}{b},\ \dfrac{1}{c}) are in A.P.
From the previous result we have ( b = \dfrac{2ac}{a+c} ).
Common difference: [ d = \dfrac{1}{b} - \dfrac{1}{a} = \dfrac{a-b}{ab} ]
Substitute ( b ): [ a - b = a - \dfrac{2ac}{a+c} = \dfrac{a(a+c) - 2ac}{a+c} = \dfrac{a^2 + ac - 2ac}{a+c} = \dfrac{a^2 - ac}{a+c} = \dfrac{a(a-c)}{a+c} ]
[ ab = a\cdot\dfrac{2ac}{a+c} = \dfrac{2a^2 c}{a+c} ]
[ d = \dfrac{\dfrac{a(a-c)}{a+c}}{\dfrac{2a^2 c}{a+c}} = \dfrac{a(a-c)}{a+c} \cdot \dfrac{a+c}{2a^2 c} = \dfrac{a-c}{2ac} ]
Proved.