1. Given condition

The arithmetic mean of $a$ and $b$ is five times their geometric mean:

$\frac{a+b}{2} = 5\sqrt{ab}$

So,

$a+b = 10\sqrt{ab}$

2. Use the identity involving $(a-b)^2$

We know:

$(a-b)^2 = (a+b)^2 - 4ab$

Now,

$a+b = 10\sqrt{ab} \implies (a+b)^2 = 100ab$

Hence,

$(a-b)^2 = 100ab - 4ab = 96ab$

Therefore,

$a-b = 4\sqrt{6ab}$

Since $a>b>0$, we take the positive root.

3. Find the required ratio

We need

$\frac{a+b}{a-b}$

Substitute:

$\frac{a+b}{a-b} = \frac{10\sqrt{ab}}{4\sqrt{6ab}}$

Cancel $\sqrt{ab}$:

$\frac{a+b}{a-b} = \frac{10}{4\sqrt6} = \frac{5}{2\sqrt6}$

Rationalizing:

$\frac{5}{2\sqrt6} \cdot \frac{\sqrt6}{\sqrt6} = \frac{5\sqrt6}{12}$

4. Match with options

$\frac{a+b}{a-b} = \frac{5\sqrt6}{12}$

So the correct option is D.