(iii) (a+b+c)² / (a² + b² + c²) = (a+b+c) / (a-b+c)

(iv) 1/(a² – b²) + 1/b² = 1/(b² – c²)

Solution:

(iii) (a+b+c)² / (a² + b² + c²) = (a+b+c) / (a-b+c)

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

Making use of the property of geometric mean, we can write:

b² = ac

Let us take the LHS: (a+b+c)² / (a² + b² + c²)

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

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

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

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

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

$ =\text{ }\left( a+b+c \right)\text{ }/\text{ }\left( a-b+c \right)=\text{ }RHS $

∴ LHS = RHS

Hence proved.

(iv) 1/(a² – b²) + 1/b² = 1/(b² – c²)

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

Making use of the property of geometric mean, we can write:

b² = ac

Let us consider LHS: 1/(a² – b²) + 1/b²

Let us take LCM

$ =1/\left( {{a}^{2}}-{{b}^{2}} \right)\text{ }+\text{ }1/{{b}^{2}}~ $

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

$ =\text{ }{{a}^{2~}}/\text{ }\left( {{a}^{2}}{{b}^{2}}-{{b}^{4}} \right) $

$ =\text{ }{{a}^{2}}~/\text{ }\left( {{a}^{2}}{{b}^{2}}-{{\left( {{b}^{2}} \right)}^{2}} \right) $

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

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

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

$ =\text{ }1/\text{ }\left( {{b}^{2}}-{{c}^{2}} \right) $

$ =\text{ }RHS $

∴ LHS = RHS

Hence proved.