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Common Difference

Show that a-b...

Question

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

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Solution

(a – b)² , ( a² + b² ) and ( a + b )² are in A.P.

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Similar questions

\frac{\sqrt{a^{2} - b^{2}} + a}{\sqrt{a^{2} + b^{2}} + b} \div \frac{\sqrt{a^{2} + b^{2}} - b}{a - \sqrt{a^{2} - b^{2}}} = \frac{b^{2}}{d^{2}}

{(a - b)}^{2}, (a^{2} + b^{2})

{(a + b)}^{2}

(a_{1}, b_{1})

(a_{2}, b_{2})

(a_{1} + a_{2}, b_{1} + b_{2})

a_{1} b_{2} = a_{2} b_{1}

a, b, c

A . P

a^{2}, b^{2}, c^{2}

G . P

a^{2}, b^{2}, c^{2}

A . P .

b + c, c + a, a + b

H . P .

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