If a, b, c are in GP, prove that (a 2 + b 2 ), (ab + bc), (b 2 + c 2 ) are in GP.

Question

Philoid · Accepted Answer

To prove: (a² + b²), (ab + bc), (b² + c²) are in GP

Given: a, b, c are in GP

Formula used: When a,b,c are in GP, b² = ac

Proof: When a,b,c are in GP,

b² = ac … (i)

Considering (a² + b²), (ab + bc), (b² + c²)

(ab + bc)² = (a²b² + 2ab²c + b²c²)

= (a²b² + ab²c + ab²c + b²c²)

= (a²b² + b⁴ + a²c² + b²c²) [From eqn. (i)]

= [b² (a² + b²)+ c² (a² + b²)]

(ab + bc)² = [(b² + c²) (a² + b²)]

From the above equation we can say that (a² + b²), (ab + bc), (b² + c²) are in GP