If the arithmetic mean of two numbers a and b, ab0, is five times their geometric mean, …

Question

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

Accepted Answer

$\dfrac{a+b}{2}=5\sqrt{ab}\ \Rightarrow\ a+b=10\sqrt{ab}$. Let $t=\sqrt{a/b}>1$; dividing by $b$: $t^2+1=10t$, i.e. $t^2-10t+1=0$, so $t=5+2\sqrt6$ (taking $t>1$).

Then $\dfrac{a}{b}=t^2=49+20\sqrt6=\dfrac{10+4\sqrt6}{10-4\sqrt6}$ (check: this ratio equals $\dfrac{(10+4\sqrt6)^2}{100-96}=\dfrac{196+80\sqrt6}{4}=49+20\sqrt6$ ).

By componendo--dividendo, with $a:b=(10+4\sqrt6):(10-4\sqrt6)$,
$$\frac{a+b}{a-b}=\frac{20}{8\sqrt6}=\frac{5}{2\sqrt6}=\frac{5\sqrt6}{12}.$$

Solutions · 1