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

Let the two positive numbers be denoted by $$a$$ and $$b$$ with $$a > b > 0$$.

We are told that the arithmetic mean is five times the geometric mean.

The formula for the arithmetic mean (A.M.) of two numbers is $$\dfrac{a+b}{2}$$, while the formula for the geometric mean (G.M.) is $$\sqrt{ab}$$.

Writing the given condition in symbols, we have

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

To clear the denominator, we multiply both sides by $$2$$:

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

Now we square both sides so that the square root disappears:

$$\bigl(a+b\bigr)^2 = \bigl(10\sqrt{ab}\bigr)^2.$$

This gives

$$a^2 + 2ab + b^2 = 100ab.$$

Bringing every term to the left side, we obtain

$$a^2 - 98ab + b^2 = 0.$$

Because both $$a$$ and $$b$$ are positive, it is convenient to set a ratio. Let us write

$$a = kb,$$

where $$k>1$$ (since $$a > b$$). Substituting $$a = kb$$ into the quadratic equation, we get

$$\bigl(kb\bigr)^2 - 98\bigl(kb\bigr)b + b^2 = 0.$$

Each term has a factor $$b^2$$, so we divide by $$b^2$$ (remembering $$b\neq0$$):

$$k^2 - 98k + 1 = 0.$$

Now we solve this quadratic in $$k$$ using the quadratic formula

$$k = \frac{98 \pm \sqrt{98^2 - 4\cdot1\cdot1}}{2}.$$

First, compute the discriminant:

$$\Delta = 98^2 - 4 = 9604 - 4 = 9600.$$

Since $$9600 = 96 \times 100$$, we note

$$\sqrt{9600} = 10\sqrt{96} = 10 \times 4\sqrt{6} = 40\sqrt{6}.$$

Hence

$$k = \frac{98 \pm 40\sqrt{6}}{2} = 49 \pm 20\sqrt{6}.$$

Because $$k$$ must exceed $$1$$, the negative sign choice $$49 - 20\sqrt{6}$$ (which is approximately $$0.02$$) is inadmissible. Thus

$$k = 49 + 20\sqrt{6}.$$

Remembering $$k = \dfrac{a}{b}$$, we now compute the desired ratio

$$\frac{a+b}{a-b} = \frac{kb+b}{kb-b} = \frac{k+1}{k-1}.$$

Substitute $$k = 49 + 20\sqrt{6}$$:

$$\frac{k+1}{k-1} = \frac{49 + 20\sqrt{6} + 1}{49 + 20\sqrt{6} - 1} = \frac{50 + 20\sqrt{6}}{48 + 20\sqrt{6}}.$$

Factor $$2$$ from numerator and denominator to simplify:

$$\frac{50 + 20\sqrt{6}}{48 + 20\sqrt{6}} = \frac{2\bigl(25 + 10\sqrt{6}\bigr)}{2\bigl(24 + 10\sqrt{6}\bigr)} = \frac{25 + 10\sqrt{6}}{24 + 10\sqrt{6}}.$$

To remove the surd from the denominator, we multiply both numerator and denominator by the conjugate $$24 - 10\sqrt{6}$$:

$$\frac{25 + 10\sqrt{6}}{24 + 10\sqrt{6}} \times \frac{24 - 10\sqrt{6}}{24 - 10\sqrt{6}} = \frac{(25 + 10\sqrt{6})(24 - 10\sqrt{6})}{(24 + 10\sqrt{6})(24 - 10\sqrt{6})}.$$

Let us evaluate the numerator first:

$$(25)(24) = 600,$$ $$(25)(-10\sqrt{6}) = -250\sqrt{6},$$ $$(10\sqrt{6})(24) = 240\sqrt{6},$$ $$(10\sqrt{6})(-10\sqrt{6}) = -100\cdot6 = -600.$$

Adding these four results: $$600 - 600 - 250\sqrt{6} + 240\sqrt{6} = -10\sqrt{6}.$$

Next, the denominator is a difference of squares:

$$(24)^2 - (10\sqrt{6})^2 = 576 - 100\cdot6 = 576 - 600 = -24.$$

Therefore the entire fraction simplifies to

$$\frac{-10\sqrt{6}}{-24} = \frac{10\sqrt{6}}{24} = \frac{5\sqrt{6}}{12}.$$

Thus we have shown that

$$\frac{a+b}{a-b}= \frac{5\sqrt{6}}{12}.$$

Among the given choices, this matches Option D.

Hence, the correct answer is Option D.