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

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(D) Given that the arithmetic mean $(A.M.)$ is five times the geometric mean $(G.M.)$:$\frac{a + b}{2} = 5\sqrt{ab}$Divide both sides by $\sqrt{ab}$:$

Vedclass · Accepted Answer

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

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(D) Given that the arithmetic mean $(A.M.)$ is five times the geometric mean $(G.M.)$:$\frac{a + b}{2} = 5\sqrt{ab}$Divide both sides by $\sqrt{ab}$:$\frac{a + b}{\sqrt{ab}} = 10$Let $x = \sqrt{\frac{a}{b}}$. Then $\frac{a}{b} = x^2$. The equation becomes:$\frac{x^2 + 1}{x} = 10 \implies x^2 - 10x + 1 = 0$Solving for $x$ using the quadratic formula:$x = \frac{10 \pm \sqrt{100 - 4}}{2} = 5 \pm \sqrt{24} = 5 \pm 2\sqrt{6}$Since $a > b$,$x > 1$,so $x = 5 + 2\sqrt{6}$.Then $\frac{a}{b} = (5 + 2\sqrt{6})^2 = 25 + 24 + 20\sqrt{6} = 49 + 20\sqrt{6}$.Using Componendo and Dividendo on $\frac{a}{b} = \frac{49 + 20\sqrt{6}}{1}$:$\frac{a + b}{a - b} = \frac{49 + 20\sqrt{6} + 1}{49 + 20\sqrt{6} - 1} = \frac{50 + 20\sqrt{6}}{48 + 20\sqrt{6}} = \frac{25 + 10\sqrt{6}}{24 + 10\sqrt{6}}$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 $b$,where $a > b > 0$,is five times their geometric mean,then $\frac{a + b}{a - b}$ is equal to

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