For the two positive numbers a , b , if a , b and 1 18 are in a geometric progression, while 1 a , 10 and 1…

Since, $a,b,\frac{1}{18}$ are in GP, so

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

$\Rightarrow a=18b^2 ....\left(1\right)$

Also, $\frac{1}{a},10,\frac{1}{ b}$ are in A.P., so

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

$\Rightarrow a+b=20ab$

$\Rightarrow 18b^2+b=360b^3$

$\Rightarrow 360b^2-18b-1=0 \left[\because b\ne 0\right]$

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

$\Rightarrow b=\frac{18+\sqrt{1764}}{720} \left\{\because b>0\right\}$

$\Rightarrow b=\frac{1}{12}$

$\Rightarrow a=18\times \frac{1}{144}=\frac{1}{8}$

Now, $16a+12b=16\times \frac{1}{8}+12\times \frac{1}{12}=3$