Show that ${{\left( a-b \right)}^{2}}$, $\left( {{a}^{2}}+{{b}^{2}} \right)$ and ${{\left( a+b \right)}^{2}}$ are in A.P.

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Show that ${{\left( a-b \right)}^{2}}$ , $\left( {{a}^{2}}+{{b}^{2}} \right)$ and ${{\left( a+b \right)}^{2}}$ are in A.P.

Show that ${{\left( a-b \right)}^{2}}$ , $\left( {{a}^{2}}+{{b}^{2}} \right)$ and ${{\left( a+b \right)}^{2}}$ are in A.P.

If we have ${{\left( a-b \right)}^{2}}$ , $\left( {{a}^{2}}+{{b}^{2}} \right)$ and ${{\left( a+b \right)}^{2}}$ to be in A.P. then,

Then It should satisfy the condition,

$2b=a+c$ [for a, b, c are in A.P]

Therefore,

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

$2\left( {{a}^{2}}+{{b}^{2}} \right)={{a}^{2}}+{{b}^{2}}-2ab+{{a}^{2}}+{{b}^{2}}+2ab$

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

LHS = RHS

Hence proved.

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