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

(a² − b²)(c² − d²) = (b² − c²)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 (a² − b²)(c² − d²) = (b² − c²)², we get :

⇒ (a² - b²)(c² - d²)

⇒ [(dk³)² - (dk²)²] [(dk)² - d²]

⇒ [d²k⁶ - d²k⁴] [d²k² - d²]

⇒ d⁴(k⁶ - k⁴)(k² - 1)

⇒ d⁴ k⁴(k² - 1)(k² - 1)

⇒ d⁴ k⁴ (k² - 1)².

Substituting values of a, b and c in R.H.S. of (a² − b²)(c² − d²) = (b² − c²)², we get :

⇒ (b² - c²)²

⇒ [(dk²)² - (dk)²]²

⇒ [d²k⁴ - d²k²]²

⇒ [d²k²(k² - 1)]²

⇒ d⁴ k⁴ (k² - 1)².

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

Hence, proved that (a² − b²)(c² − d²) = (b² − c²)² .