Show that $(a−b)^2,\ (a^2+b^2),\ (a+b)^2$ are in A.P.
Given: $(a−b)^2,\ (a^2+b^2),\ (a+b)^2$
To do: To show that $(a−b)^2,\ (a^2+b^2),\ (a+b)^2$ are in A.P.
Solution:
Assume that $( a−b)^2,\ ( a^2+b^2)$ and $(a+b)^2$ are in AP.
$( a^2+b^2)−( a−b)^2=( a+b)^2−( a^2+b^2)$
$(a^2+b^2)−(a^2+b^2−2ab)=a^2+b^2+2ab−a^2−b^2$
$2ab=2ab$
Because, difference between two consecutive terms is the same.
Hence given terms are in A.P.
 
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