Show that $(a – b)^2, (a^2 + b^2)$ and $(a + b)^2$ are in A.P.

Given:

Given terms are $(a−b)^2,\ (a^2+b^2),\ (a+b)^2$. 

To do:

We have to show that $(a−b)^2,\ (a^2+b^2),\ (a+b)^2$ are in A.P.

Solution:

If the given terms are in A.P., then their common difference should be equal.

$( a^2+b^2)−( a−b)^2=(a^2+b^2)−(a^2+b^2−2ab)$

$=2ab$

$( a+b)^2−( a^2+b^2)=a^2+b^2+2ab−a^2−b^2$

$=2ab$

Therefore,

$( a^2+b^2)−( a−b)^2=( a+b)^2−( a^2+b^2)$

Hence, the given terms are in A.P.