The arithmetic mean between $a$ and $b$ is twice the geometric mean between $a$ and $b$ .
Prove that $: \frac{a}{b} = 7 + 4 \sqrt{3}$ or $7 - 4 \sqrt{3}$ .

Open in App

Solution

Given that
A.M of $, b = 2 \times$ G.M of $a, b$

$\Rightarrow$ $\frac{a + b}{2} = 2 \sqrt{a b}$

$\Rightarrow$ $a + b = 4 \sqrt{a b}$

Squaring both sides, we get

$\Rightarrow$ $a^{2} + b^{2} + 2 a b = 16 a b$

$\Rightarrow$ $a^{2} + b^{2} = 14 a b$

$\Rightarrow$ $\frac{a}{b} + \frac{b}{a} = 14$

Let $\frac{a}{b} = x$

$\Rightarrow$ $x + \frac{1}{x} = 14$

$\Rightarrow$ $x^{2} - 14 x + 1 = 0$

$\Rightarrow$ $x = \frac{- (- 14) \pm \sqrt{14^{2} - 4}}{2}$

$\Rightarrow$ $\frac{14 \pm \sqrt{192}}{2}$

$∴$ $x = \frac{a}{b} = 7 + 4 \sqrt{3}$ or $7 - 4 \sqrt{3}$

Suggest Corrections

Similar questions

a

b

a : b = 2 + \sqrt{3} : 2 - \sqrt{3}

If A and G are respectively the arithmetic and geometric means between two distinct positive numbers a and b then prove that A>G.

b

a

x; b

a^{'}

y; b

(a, b, x, y, z > 0)

x y z

a : b = (2 + \sqrt{3}) : (2 - \sqrt{3}) .

Which of the following is the correct relation between a (arithmetic mean), b (geometric mean) and c (harmonic mean)?

Join BYJU'S Learning Program