If a, b, c are in continued proportion and a(b - c) = 2b, prove that :

a - c = $\dfrac{2(a + b)}{a}$

Since, a, b and c are in continued proportion,

$\therefore \dfrac{a}{b} = \dfrac{b}{c}$

⇒ b² = ac ….(i)

Given,

⇒ a(b - c) = 2b

⇒ ab - ac = 2b

⇒ ab - b² = 2b [From eq (i)]

⇒ b(a - b) = 2b

⇒ a - b = 2 ….(ii)

To prove : a - c = $\dfrac{2(a + b)}{a}$.

Considering, L.H.S.

$= a - c \\[1em] = \dfrac{a(a - c)}{a} \\[1em] = \dfrac{a^2 - ac}{a} \\[1em] = \dfrac{a^2 - b^2}{a} \\[1em] = \dfrac{(a - b)(a + b)}{a} \\[1em] = \dfrac{2(a + b)}{a} \text{ [From eq (ii)]}$

As, L.H.S. = $\dfrac{2(a + b)}{a}$ = R.H.S.

Hence, proved that a - c = $\dfrac{2(a + b)}{a}$.