Question

If a, b, c are in GP, prove that (a² + b² ), (ab + bc), (b² + c² ) are in GP.

Answer 1

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