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(C) Let the two numbers be $a$ and $b$.Given that $A = \frac{a+b}{2}$ and $G = \sqrt{ab}$.From these,we have $a+b = 2A$ and $ab = G^2$.We know that $(

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

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(C) Let the two numbers be $a$ and $b$.Given that $A = \frac{a+b}{2}$ and $G = \sqrt{ab}$.From these,we have $a+b = 2A$ and $ab = G^2$.We know that $(a-b)^2 = (a+b)^2 - 4ab$.Substituting the values,$(a-b)^2 = (2A)^2 - 4G^2 = 4(A^2 - G^2)$.Therefore,$a-b = 2\sqrt{A^2 - G^2}$ (assuming $a > b$).Adding the equations $a+b = 2A$ and $a-b = 2\sqrt{A^2 - G^2}$,we get $2a = 2A + 2\sqrt{A^2 - G^2}$,which implies $a = A + \sqrt{A^2 - G^2}$.Subtracting the equations,we get $2b = 2A - 2\sqrt{A^2 - G^2}$,which implies $b = A - \sqrt{A^2 - G^2}$.Thus,the numbers are $A \pm \sqrt{A^2 - G^2}$.

If the arithmetic mean and geometric mean of two positive numbers are $A$ and $G$ respectively,then the numbers are ..........

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