Question
Show that $(a - b)^2, (a^2 + b^2)$ and $(a + b)^2$ are in AP.
✓
Answer
The given number are $(a - b)^2, (a^2 + b^2)$ and $(a + b)^2$.
Now,
$(a^2 + b^2) - (a - b)^2 = a^2 + b^2 - (a^2 - 2ab + b^2) = a^2 + b^2 - a^2 + 2ab - b^2 = 2ab$
$(a + b)^2 - (a^2+ b^2) = a^2 + 2ab + b^2 - a^2 - b^2 = 2ab$
So, $(a^2 + b^2) - (a - b)^2 = (a + b)^2 - (a^2 + b^2) = 2ab$ (Constant)
Since each term differs from its preceding term by a constant, therefore, the given numbers are in AP.
Now,
$(a^2 + b^2) - (a - b)^2 = a^2 + b^2 - (a^2 - 2ab + b^2) = a^2 + b^2 - a^2 + 2ab - b^2 = 2ab$
$(a + b)^2 - (a^2+ b^2) = a^2 + 2ab + b^2 - a^2 - b^2 = 2ab$
So, $(a^2 + b^2) - (a - b)^2 = (a + b)^2 - (a^2 + b^2) = 2ab$ (Constant)
Since each term differs from its preceding term by a constant, therefore, the given numbers are in AP.
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