( \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.
11TH•Mathematics•Sequences And Series•Question 17
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If $\dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}$ are in A.P., show that $b=\dfrac{2ac}{a+c}$.
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