Show that ${(a - b)}^{2}, (a^{2} + b^{2})$ and ${(a + b)}^{2}$ are in AP.

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Solution

The given numbers are ${(a - b)}^{2}, (a^{2} + b^{2})$ and ${(a + b)}^{2}$ .
Now,

$(a^{2} + b^{2}) - {(a - b)}^{2} = a^{2} + b^{2} - (a^{2} - 2 a b + b^{2}) = a^{2} + b^{2} - a^{2} + 2 a b - b^{2} = 2 a b$

${(a + b)}^{2} - (a^{2} + b^{2}) = a^{2} + 2 a b + b^{2} - a^{2} - b^{2} = 2 a b$

So, $(a^{2} + b^{2}) - {(a - b)}^{2} = {(a + b)}^{2} - (a^{2} + b^{2}) = 2 a b (Constant)$

Since each term differs from its preceding term by a constant, therefore, the given numbers are in AP.

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Show that $(a - b)^{2}, (a^{2} + b^{2})$ and $(a + b)^{2}$ are in A.P.

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