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

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(B) Given that the arithmetic mean $A = \frac{a+b}{2}$ and geometric mean $G = \sqrt{ab}$.According to the problem,$A = 2G$.$\frac{a+b}{2} = 2\sqrt{ab

Vedclass · Accepted Answer

$\frac{2 + \sqrt{3}}{2 - \sqrt{3}}$

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(B) Given that the arithmetic mean $A = \frac{a+b}{2}$ and geometric mean $G = \sqrt{ab}$.According to the problem,$A = 2G$.$\frac{a+b}{2} = 2\sqrt{ab} \implies \frac{a+b}{\sqrt{ab}} = 4$.Dividing by $\sqrt{ab}$,we get $\frac{a}{\sqrt{ab}} + \frac{b}{\sqrt{ab}} = 4 \implies \sqrt{\frac{a}{b}} + \sqrt{\frac{b}{a}} = 4$.Let $x = \sqrt{\frac{a}{b}}$. Then $x + \frac{1}{x} = 4 \implies x^2 - 4x + 1 = 0$.Using the quadratic formula,$x = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$.Since $x = \sqrt{\frac{a}{b}}$,we have $\frac{a}{b} = x^2 = (2 \pm \sqrt{3})^2 = 4 + 3 \pm 4\sqrt{3} = 7 \pm 4\sqrt{3}$.Alternatively,using the componendo and dividendo method on $\frac{a+b}{2\sqrt{ab}} = \frac{2}{1}$:$\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}} = \frac{2+1}{2-1} = 3$.$\frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2} = 3 \implies \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} = \sqrt{3}$.Applying componendo and dividendo again: $\frac{2\sqrt{a}}{2\sqrt{b}} = \frac{\sqrt{3}+1}{\sqrt{3}-1}$.$\frac{a}{b} = \left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)^2 = \frac{3+1+2\sqrt{3}}{3+1-2\sqrt{3}} = \frac{4+2\sqrt{3}}{4-2\sqrt{3}} = \frac{2+\sqrt{3}}{2-\sqrt{3}}$.

If the arithmetic mean of two numbers $a$ and $b$ is twice their geometric mean,then $a : b = \dots$

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