For the two positive numbers a, b , if a, b and (1)/(18) are in a...

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

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

$⇒ a = 18{b}^{2} . . . . \left(1\right)$

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

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

$⇒ a + b = 20ab$

$⇒ 18{b}^{2} + b = 360{b}^{3}$

$⇒ 360{b}^{2} - 18b - 1 = 0\left[∵ b \ne 0\right]$

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

$⇒ b = \dfrac{18 + \sqrt{1764}}{720}\left\{∵ b > 0\right\}$

$⇒ b = \dfrac{1}{12}$

$⇒ a = 18 \times \dfrac{1}{144} = \dfrac{1}{8}$

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