If a ≠ b and a : b is the duplicate ratio of (a + c) and (b + c), show that a, c and b are in continued proportion.

Given,

a : b is the duplicate ratio of (a + c) and (b + c).

$\therefore \dfrac{a}{b} = \dfrac{a + c}{b + c} \times \dfrac{a + c}{b + c} \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{a^2 + ac + ac + c^2}{b^2 + bc + bc + c^2} \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{a^2 + 2ac + c^2}{b^2 + 2bc + c^2} \\[1em] \Rightarrow a(b^2 + 2bc + c^2) = b(a^2 + 2ac + c^2) \\[1em] \Rightarrow ab^2 + 2abc + ac^2 = ba^2 + 2abc + bc^2 \\[1em] \Rightarrow ac^2 - bc^2 = ba^2 - ab^2 + 2abc - 2abc \\[1em] \Rightarrow c^2(a - b) = ab(a - b) \\[1em] \Rightarrow c^2 = ab.$

Since, c² = ab.

Hence, proved that a, c and b are in continued proportion.