If a, b, c, d are in continued proportion, prove that :

(a + b)(b + c) - (a + c)(b + d) = (b - c)²

Given a, b, c, d are in continued proportion.

∴ a : b = b : c = c : d

$\dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{d}$ = k (let)

⇒ c = dk, b = ck = (dk)k = dk², a = bk = (dk²)k = dk³.

Substituting values of a, b and c in L.H.S. of (a + b)(b + c) - (a + c)(b + d) = (b - c)², we get :

⇒ (dk³ + d)(dk² + dk) - (dk³ + dk)(dk² + d)

⇒ d(k³ + 1) dk(k + 1) - dk(k² + 1) d(k² + 1)

⇒ d²k(k³ + 1)(k + 1) - d²k(k² + 1)(k² + 1)

⇒ d²k[(k⁴ + k³ + k + 1) - (k⁴ + 2k² + 1)]

⇒ d²k[k⁴ + k³ + k + 1 - k⁴ - 2k² - 1]

⇒ d²k[k³ - 2k² + k]

⇒ d²k²[k² - 2k + 1]

⇒ d²k²(k - 1)²

Substituting values of a, b and c in R.H.S. of (a + b)(b + c) - (a + c)(b + d) = (b - c)², we get :

⇒ (b - c)²

⇒ (dk² - dk)²

⇒ (dk[k - 1])²

⇒ d²k²(k - 1)²

Since, L.H.S. = R.H.S.

Hence, (a + b)(b + c) - (a + c)(b + d) = (b - c)².