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