If a, b and c are in G.P. and a, x, b, y, c are in A.P. prove that :

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

(ii) $\dfrac{a}{x} + \dfrac{c}{y} = 2.$

Given,

a, b and c are in G.P.

⇒ b² = ac ………….(i)

a, x, b, y, c are in A.P.

⇒ 2x = a + b and 2y = b + c.

⇒ $x = \dfrac{a + b}{2} \text{ and } y = \dfrac{b + c}{2}$ ……..(ii)

(i) Substituting value of x and y from (ii) in L.H.S. of $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{2}{b}$,

$\text{L.H.S}. = \dfrac{1}{\dfrac{a + b}{2}} + \dfrac{1}{\dfrac{b + c}{2}} \\[1em] = \dfrac{2}{a + b} + \dfrac{2}{b + c} \\[1em] = \dfrac{2(b + c) + 2(a + b)}{(a + b)(b + c)} \\[1em] = \dfrac{2b + 2c + 2a + 2b}{(a + b)(b + c)}\\[1em] = \dfrac{2a + 2c + 4b}{ab + ac + b^2 + bc} \\[1em] = \dfrac{2(a + c + 2b)}{ab + b^2 + b^2 + bc} \text{from (i)} \\[1em] = \dfrac{2(a + c + 2b)}{b(a + b + b + c)} \\[1em] = \dfrac{2(a + c + 2b)}{b(a + c + 2b)} \\[1em] = \dfrac{2}{b} = \text{R.H.S.}$

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

(ii) Substituting value of x and y from (ii) in L.H.S. of $\dfrac{a}{x} + \dfrac{c}{y} = 2$,

$\text{L.H.S.} = \dfrac{a}{\dfrac{a + b}{2}} + \dfrac{c}{\dfrac{b + c}{2}} \\[1em] = \dfrac{2a}{a + b} + \dfrac{2c}{b + c} \\[1em] = 2\Big(\dfrac{a}{a + b} + \dfrac{c}{b + c}\Big) \\[1em] = 2\Big[\dfrac{a(b + c) + c(a + b)}{(a + b)(b + c)}\Big] \\[1em] = 2\Big(\dfrac{ab + ac + ac + bc}{ab + ac + b^2 + bc}\Big) \\[1em] = 2\Big(\dfrac{ab + b^2 + b^2 + bc}{ab + b^2 + b^2 + bc}\Big) ……\text{from (i)} \\[1em] = 2 = \text{R.H.S.}$

Hence, proved that $\dfrac{a}{x} + \dfrac{c}{y} = 2$.