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

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If A and G be A.M. and G.M., respectively between two positive numbers,
Two numbers be a and b.

$AM=A=\frac{a+b}{2}$

$\Rightarrow a+b=2A$ ...................................................................1

$GM=G=\sqrt{ab}$

$\Rightarrow ab=G^2$ ...........................................................................2

We know $(a-b)^2=(a+b)^2-4ab$

Put values from equation 1 and 2,

$(a-b)^2=4A^2-4G^2$

$(a-b)^2=4(A^2-G^2)$

$(a-b)^2=4(A+G)(A-G)$

$(a-b)=4\sqrt{(A+G)(A-G)}$ ..................................................................3

From 1 and 3 , we have

$2a=2A+2\sqrt{(A+G)(A-G)}$

$\Rightarrow a=A+\sqrt{(A+G)(A-G)}$

Put value of a in equation 1, we get

$b=2A-A-\sqrt{(A+G)(A-G)}$

$\Rightarrow b=A-\sqrt{(A+G)(A-G)}$

Thus, numbers are $A \pm \sqrt{( A+G)(A-G)}$

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