If $A$ and $G$ be the $A . M$ and $G . M .$ between two numbers, prove that the numbers are $A \pm \sqrt{(A + G) (A - G)} .$

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Solution

Let the numbers be $a$ and $b$
then $A = A . M . = \frac{a + b}{2}$ or $a + b = 2 A$
$G = G . M . = \sqrt{a b}$ $∴ a b = G^{2}$
From $(1)$ and $(2)$ we find that $a$ and $b$ are the roots of $t^{2} - 2 A t + G^{2} = 0$
$∴ t = \frac{2 A \pm \sqrt{4 A^{2} - {4 G}^{2}}}{2} = A \pm \sqrt{(A - G) (A + G)}$

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Similar questions

A \pm \sqrt{(A + G) (A - G)} .

If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are $A \pm \sqrt{(A + G) (A - G)} .$

a

b

c

G_{1}, G_{2}

G_{1}^{3} + G_{2}^{3} = 2 a b c

A \pm \sqrt{A^{2} - G^{2}}

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