(iii) a² + b², ab + bc, b² + c²

Solution:

(iii) a² + b², ab + bc, b² + c²

According to the question, a, b, c are in GP.

Making use of 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 take the LHS: (ab + bc)²

$ {{\left( ab\text{ }+\text{ }bc \right)}^{2}}~=\text{ }{{a}^{2}}{{b}^{2}}~+\text{ }2a{{b}^{2}}c\text{ }+\text{ }{{b}^{2}}{{c}^{2}} $

$ =\text{ }{{a}^{2}}{{b}^{2}}~+\text{ }2{{b}^{2}}\left( {{b}^{2}} \right)\text{ }+\text{ }{{b}^{2}}{{c}^{2}}~\left[ Since,\text{ }ac\text{ }=\text{ }{{b}^{2}} \right] $

$ =\text{ }{{a}^{2}}{{b}^{2}}~+\text{ }2{{b}^{4}}~+\text{ }{{b}^{2}}{{c}^{2}} $

$ =\text{ }{{a}^{2}}{{b}^{2}}~+\text{ }{{b}^{4}}~+\text{ }{{a}^{2}}{{c}^{2}}~+\text{ }{{b}^{2}}{{c}^{2}}~\left\{ again\text{ }using\text{ }{{b}^{2}}~=\text{ }ac\text{ } \right\} $

$ =\text{ }{{b}^{2}}\left( {{b}^{2}}~+\text{ }{{a}^{2}} \right)\text{ }+\text{ }{{c}^{2}}\left( {{a}^{2}}~+\text{ }{{b}^{2}} \right) $

$ =\text{ }\left( {{a}^{2}}~+\text{ }{{b}^{2}} \right)\left( {{b}^{2}}~+\text{ }{{c}^{2}} \right) $

$ =\text{ }RHS $

∴ LHS = RHS

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