Question

if the arithmetic mean of two positive numbers a and b (a>b) is twice their geometric mean then find a:b

Answer 1

$let \: the \: numbers \: be \: a \: \: and \: \: b$
$given \: \frac{a + b}{2} = 2 \sqrt{ab}$
Squaring both sides and rearranging,
${a}^{2} - 14ab + {b}^{2} = 0$
$dividing \: through \: out \: by \: {b}^{2}$
${( \frac{a}{b} })^{2} - 14( \frac{a}{b} ) + 1 = 0$
$solved \: for \: \frac{a}{b} \: using \: standard \: formula$
$\frac{a}{b} = 7 + 4 \sqrt{3} \: \: or \: 7 - 4 \sqrt{3}$

Answer 2

a : b = 1393 : 100

Step-by-step explanation:

The arithmetic mean of two numbers a and b is $\frac{a + b}{2}$ and the geometric mean of two same numbers is $\sqrt{ab}$.

Given that, $\frac{a + b}{2} = 2 \times \sqrt{ab}$

Now, squaring both sides we get,

(a + b)² = 16ab

⇒ a² + 2ab + b² = 16ab

⇒ a² - 14ab + b² = 0

Dividing both sides by b², we get

$(\frac{a}{b})^{2} - 14 (\frac{a}{b}) + 1 = 0$

⇒ $(\frac{a}{b}) = \frac{-(-14) \pm\sqrt{(- 14)^{2} - 4 (1)(1)} }{2(1)}$

⇒ $(\frac{a}{b}) = 13.93$ {Since a > b} (Approx.)

⇒ a : b = 1393 : 100 (Answer)