For the two positive numbers $a,b,$ if $a,b$ and $\frac{1}{18}$ are in a geometric progression, while $\frac{1}{a},10$ and $\frac{1}{b}$ are in an arithmetic progression, then $16a+12b$ is equal to _________.

$$ \begin{aligned} & \mathrm{a}, \mathrm{b}, \frac{1}{18} \rightarrow \mathrm{GP} \\\\ & \frac{\mathrm{a}}{18}=\mathrm{b}^2\quad...(i) \\\\ & \frac{1}{\mathrm{a}}, 10, \frac{1}{\mathrm{~b}} \rightarrow \mathrm{AP} \\\\ & \frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}=20 \\\\ & \Rightarrow \mathrm{a}+\mathrm{b}=20 \mathrm{ab}, \text { from eq. (i) } ; \text { we get } \\\\ & \Rightarrow 18 \mathrm{~b}^2+\mathrm{b}=360 \mathrm{~b}^3 \\\\ & \Rightarrow 360 \mathrm{~b}^2-18 \mathrm{~b}-1=0 \quad\{\because \mathrm{b} \neq 0\} \\\\ & \Rightarrow \mathrm{b}=\frac{18 \pm \sqrt{324+1440}}{720} \\\\ & \Rightarrow b =\frac{18+\sqrt{1764}}{720} \quad\{\because \mathrm{b}>0\} \\\\ & \Rightarrow b = \frac{1}{12} \\\\ & \Rightarrow \mathrm{a}=18 \times \frac{1}{144}=\frac{1}{8} \\\\ & \text{Now},16\mathrm{a}+12 \mathrm{~b}=16 \times \frac{1}{8}+12 \times \frac{1}{12}=3 \end{aligned} $$