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.
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