Prove that $(a-b)^{2}, (a^{2}+b^{2})$ and $(a+b)^{2}$ form an $A.P.$

(N/A) To prove that the given terms $(a-b)^{2}, (a^{2}+b^{2}),$ and $(a+b)^{2}$ form an $A.P.$,we need to show that the difference between consecutive terms is constant.
Let the terms be $T_1 = (a-b)^2$,$T_2 = (a^2+b^2)$,and $T_3 = (a+b)^2$.
First,calculate the difference between the second and first term $(d_1)$:
$d_1 = T_2 - T_1 = (a^2 + b^2) - (a - b)^2$
$d_1 = (a^2 + b^2) - (a^2 - 2ab + b^2)$
$d_1 = a^2 + b^2 - a^2 + 2ab - b^2 = 2ab$
Next,calculate the difference between the third and second term $(d_2)$:
$d_2 = T_3 - T_2 = (a + b)^2 - (a^2 + b^2)$
$d_2 = (a^2 + 2ab + b^2) - (a^2 + b^2)$
$d_2 = a^2 + 2ab + b^2 - a^2 - b^2 = 2ab$
Since $d_1 = d_2 = 2ab$,the common difference is constant.
Therefore,the given terms form an $A.P.$