Let $$a_1, a_2, \ldots, a_n$$ be in A.P. If $$a_5 = 2a_7$$ and $$a_{11} = 18$$, then $$12\left(\dfrac{1}{\sqrt{a_{10}} + \sqrt{a_{11}}} + \dfrac{1}{\sqrt{a_{11}} + \sqrt{a_{12}}} + \ldots + \dfrac{1}{\sqrt{a_{17}} + \sqrt{a_{18}}}\right)$$ is equal to ______.

Given $$a_5 = 2a_7$$ and $$a_{11} = 18$$.

$$a + 4d = 2(a + 6d) \implies a = -8d$$

$$a_{11} = a + 10d = -8d + 10d = 2d = 18 \implies d = 9, \; a = -72$$

Each term in the sum can be rationalized:

$$\frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} = \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{a_{k+1} - a_k} = \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d} = \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{9}$$

The sum telescopes:

$$\sum_{k=10}^{17}\frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} = \frac{\sqrt{a_{18}} - \sqrt{a_{10}}}{9}$$

$$a_{10} = -72 + 81 = 9, \quad a_{18} = -72 + 153 = 81$$

$$= \frac{9 - 3}{9} = \frac{2}{3}$$

Therefore: $$12 \times \frac{2}{3} = 8$$