if a, b, c are in G.P., prove that (a + 2b + 2c) (a – 2b + 2c) = a2 + 4c2

(a + 2b + 2c) (a – 2b + 2c) = a² + 4c²

Given that a, b, c are in GP.

By using the property of geometric mean,

b² = ac

Let us consider LHS: (a + 2b + 2c) (a – 2b + 2c)

Upon expansion we get,

(a + 2b + 2c) (a – 2b + 2c) = a² – 2ab + 2ac + 2ab – 4b² + 4bc + 2ac – 4bc + 4c²

= a² + 4ac – 4b² + 4c²

= a² + 4ac – 4(ac) + 4c² [Since, b² = ac]

= a² + 4c²

= RHS

LHS = RHS

Hence proved.