If a, b, c are in G.P; a, x, b are in A.P. and b, y, c are also in A.P.

Prove that : $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{2}{b}$.

Given,

a, x, b are in A.P.

∴ b - x = x - a

⇒ 2x = a + b

⇒ x = $\dfrac{a + b}{2}$ ………(1)

Given,

b, y, c are in A.P.

∴ c - y = y - b

⇒ 2y = b + c

⇒ y = $\dfrac{b + c}{2}$ ………..(2)

Given,

a, b and c are in G.P.

⇒ $\dfrac{b}{a} = \dfrac{c}{b}$

⇒ b² = ac ……….(3)

To prove :

$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{2}{b}$

Substituting value of x and y from equation (1) and (2), in L.H.S. of above equation :

$\Rightarrow \dfrac{1}{\dfrac{a + b}{2}} + \dfrac{1}{\dfrac{b + c}{2}} \\[1em] \Rightarrow \dfrac{2}{a + b} + \dfrac{2}{b + c} \\[1em] \Rightarrow \dfrac{2(b + c) + 2(a + b)}{(a + b)(b + c)} \\[1em] \Rightarrow \dfrac{2b + 2c + 2a + 2b}{ab + ac + b^2 + bc} \\[1em] \Rightarrow \dfrac{2a + 2c + 4b}{ab + b^2 + b^2 + bc} \space [\text{From} (3)] \\[1em] \Rightarrow \dfrac{2(a + c + 2b)}{ab + 2b^2 + bc} \\[1em] \Rightarrow \dfrac{2(a + c + 2b)}{b(a + 2b + c)} \\[1em] \Rightarrow \dfrac{2}{b}.$

Hence, proved that $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{2}{b}$.