Question:
Show that $(a-b)^{2},\left(a^{2}+b^{2}\right)$ and $(a+b)^{2}$ are in AP.
Solution:
The given numbers are $(a-b)^{2},\left(a^{2}+b^{2}\right)$ and $(a+b)^{2}$.
Now,
$\left(a^{2}+b^{2}\right)-(a-b)^{2}=a^{2}+b^{2}-\left(a^{2}-2 a b+b^{2}\right)=a^{2}+b^{2}-a^{2}+2 a b-b^{2}=2 a b$
$(a+b)^{2}-\left(a^{2}+b^{2}\right)=a^{2}+2 a b+b^{2}-a^{2}-b^{2}=2 a b$
So, $\left(a^{2}+b^{2}\right)-(a-b)^{2}=(a+b)^{2}-\left(a^{2}+b^{2}\right)=2 a b$ (Constant)
Since each term differs from its preceding term by a constant, therefore, the given numbers are in AP.
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