If the arithmetic mean of two numbers a and b | vedclass

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(D) Given that the arithmetic mean $(AM)$ of $a$ and $b$ is five times their geometric mean $(GM)$:$\frac{a + b}{2} = 5\sqrt{ab}$$\frac{a + b}{\sqrt{a

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$\frac{5\sqrt{6}}{12}$

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(D) Given that the arithmetic mean $(AM)$ of $a$ and $b$ is five times their geometric mean $(GM)$:$\frac{a + b}{2} = 5\sqrt{ab}$$\frac{a + b}{\sqrt{ab}} = 10$Let $x = \sqrt{\frac{a}{b}}$. Then $\frac{a+b}{\sqrt{ab}} = \frac{a}{\sqrt{ab}} + \frac{b}{\sqrt{ab}} = \sqrt{\frac{a}{b}} + \sqrt{\frac{b}{a}} = x + \frac{1}{x} = 10$.Solving $x^2 - 10x + 1 = 0$,we get $x = \frac{10 \pm \sqrt{100 - 4}}{2} = 5 \pm \sqrt{24} = 5 \pm 2\sqrt{6}$.Since $a > b$,$x > 1$,so $x = 5 + 2\sqrt{6}$.We need to find $\frac{a+b}{a-b} = \frac{\frac{a}{b} + 1}{\frac{a}{b} - 1} = \frac{x^2 + 1}{x^2 - 1}$.Since $x = 5 + 2\sqrt{6}$,$x^2 = (5 + 2\sqrt{6})^2 = 25 + 24 + 20\sqrt{6} = 49 + 20\sqrt{6}$.$\frac{a+b}{a-b} = \frac{49 + 20\sqrt{6} + 1}{49 + 20\sqrt{6} - 1} = \frac{50 + 20\sqrt{6}}{48 + 20\sqrt{6}} = \frac{25 + 10\sqrt{6}}{24 + 10\sqrt{6}}$.Rationalizing the denominator:$\frac{(25 + 10\sqrt{6})(24 - 10\sqrt{6})}{(24)^2 - (10\sqrt{6})^2} = \frac{600 - 250\sqrt{6} + 240\sqrt{6} - 600}{576 - 600} = \frac{-10\sqrt{6}}{-24} = \frac{5\sqrt{6}}{12}$.

If the arithmetic mean of two numbers $a$ and $b$,$a > b > 0$,is five times their geometric mean,then $\frac{a + b}{a - b}$ is equal to

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