1. Use the geometric progression condition

If $a,b,\dfrac{1}{18}$ are in G.P., then the middle term squared equals the product of the other two terms:

$b^2=a\cdot \frac{1}{18}$

So,

$a=18b^2$

2. Use the arithmetic progression condition

If $\dfrac{1}{a},10,\dfrac{1}{b}$ are in A.P., then the middle term is the average of the other two:

$10=\frac{\frac{1}{a}+\frac{1}{b}}{2}$

Thus,

$\frac{1}{a}+\frac{1}{b}=20$

Now substitute $a=18b^2$:

$\frac{1}{18b^2}+\frac{1}{b}=20$

Multiply by $18b^2$:

$1+18b=360b^2$

So,

$360b^2-18b-1=0$

3. Solve for $b$

Using the quadratic formula:

$b=\frac{18\pm\sqrt{(-18)^2-4(360)(-1)}}{2\cdot 360}$

$b=\frac{18\pm\sqrt{324+1440}}{720}$

$b=\frac{18\pm\sqrt{1764}}{720}$

$b=\frac{18\pm 42}{720}$

Since $b>0$,

$b=\frac{60}{720}=\frac{1}{12}$

Then,

$a=18\left(\frac{1}{12}\right)^2=18\cdot \frac{1}{144}=\frac{1}{8}$

4. Compute $16a+12b$

$16a+12b=16\left(\frac{1}{8}\right)+12\left(\frac{1}{12}\right)$

$=2+1=3$

5. Compare with stored answer

Derived answer is $3$, which matches the stored correct answer.