Question

If a, b, c are in G.P., prove that the following are also in G.P.:
(i) a², b², c²

(ii) a³, b³, c³

(iii) a² + b², ab + bc, b² + c²

Answer 1

(i) a², b², c²

Given that a, b, c are in GP.

By using the property of geometric mean,

b² = ac

on squaring both the sides we get,

(b²)² = (ac)²

(b²)² = a²c²

∴ a², b², c² are in G.P.

(ii) a³, b³, c³

Given that a, b, c are in GP.

By using the property of geometric mean,

b² = ac

on squaring both the sides we get,

(b²)³ = (ac)³

(b²)³ = a³c³

(b³)² = a³c³

∴ a³, b³, c³ are in G.P.

(iii) a² + b², ab + bc, b² + c²

Given that a, b, c are in GP.

By using the property of geometric mean,

b² = ac

a² + b², ab + bc, b² + c² or (ab + bc)² = (a² + b²) (b² + c²) [by using the property of GM]

Let us consider LHS: (ab + bc)²

Upon expansion we get,

(ab + bc)² = a²b² + 2ab²c + b²c²

= a²b² + 2b²(b²) + b²c² [Since, ac = b²]

= a²b² + 2b⁴ + b²c²

= a²b² + b⁴ + a²c² + b²c² {again using b² = ac }

= b²(b² + a²) + c²(a² + b²)

= (a² + b²)(b² + c²)

= RHS

∴ LHS = RHS

Hence a² + b², ab + bc, b² + c² are in GP.