1. Use the G.P. condition

Since $\dfrac{1}{16}, a, b$ are in G.P., the common ratio is same, so

$\frac{a}{1/16}=\frac{b}{a}$

Hence,

$a^2=\frac{1}{16}b$

or

$b=16a^2 \quad ...(1)$

2. Use the A.P. condition

Since $\dfrac{1}{a}, \dfrac{1}{b}, 6$ are in A.P., the middle term is the average of the other two:

$2\cdot \frac{1}{b}=\frac{1}{a}+6$

So,

$\frac{2}{b}=\frac{1}{a}+6 \quad ...(2)$

3. Substitute $b=16a^2$ into the A.P. equation

From (2):

$\frac{2}{16a^2}=\frac{1}{a}+6$

$\frac{1}{8a^2}=\frac{1}{a}+6$

Multiply throughout by $8a^2$:

$1=8a+48a^2$

Thus,

$48a^2+8a-1=0$

4. Solve the quadratic

$a=\frac{-8\pm\sqrt{64+192}}{96}=\frac{-8\pm16}{96}$

So,

$a=\frac{8}{96}=\frac{1}{12}$

or

$a=\frac{-24}{96}=-\frac{1}{4}$

Given $a>0$, we take

$a=\frac{1}{12}$

Now from (1):

$b=16\left(\frac{1}{12}\right)^2=16\cdot \frac{1}{144}=\frac{1}{9}$

5. Compute $72(a+b)$

$a+b=\frac{1}{12}+\frac{1}{9}=\frac{3+4}{36}=\frac{7}{36}$

Therefore,

$72(a+b)=72\cdot \frac{7}{36}=14$

Final Answer

$\boxed{14}$

The derived answer matches the stored correct answer.