Question
Show that $(a-b)^2,\left(a^2+b^2\right)$ and $(a+b)^2$ are in $AP.$
✓
Answer
The given number 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.
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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