Let $$\frac{1}{16}$$, $$a$$ and $$b$$ be in G.P. and $$\frac{1}{a}$$, $$\frac{1}{b}$$, 6 be in A.P., where $$a, b > 0$$. Then $$72(a+b)$$ is equal to ________.

Given $$\frac{1}{16}$$, $$a$$, $$b$$ are in G.P., so $$a^2 = \frac{b}{16}$$, which gives $$b = 16a^2$$.

Also $$\frac{1}{a}$$, $$\frac{1}{b}$$, $$6$$ are in A.P., so $$\frac{2}{b} = \frac{1}{a} + 6$$.

Substituting $$b = 16a^2$$: $$\frac{2}{16a^2} = \frac{1}{a} + 6$$, i.e., $$\frac{1}{8a^2} = \frac{1}{a} + 6$$.

Multiplying through by $$8a^2$$: $$1 = 8a + 48a^2$$, so $$48a^2 + 8a - 1 = 0$$.

Using the quadratic formula: $$a = \frac{-8 \pm \sqrt{64 + 192}}{96} = \frac{-8 \pm 16}{96}$$.

Since $$a > 0$$, we get $$a = \frac{-8+16}{96} = \frac{8}{96} = \frac{1}{12}$$.

Then $$b = 16 \times \frac{1}{144} = \frac{16}{144} = \frac{1}{9}$$.

Therefore $$72(a+b) = 72\left(\frac{1}{12} + \frac{1}{9}\right) = 72 \times \frac{3+4}{36} = 72 \times \frac{7}{36} = 14$$.

The answer is $$14$$.