For two positive numbers a and b,if a, b and | vedclass

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(A) Given $a, b, \frac{1}{18}$ are in $GP$,so $b^2 = a 	imes \frac{1}{18} \implies a = 18b^2$ $(i)$.Given $\frac{1}{a}, 10, \frac{1}{b}$ are in $AP$,

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(A) Given $a, b, \frac{1}{18}$ are in $GP$,so $b^2 = a 	imes \frac{1}{18} \implies a = 18b^2$ $(i)$.Given $\frac{1}{a}, 10, \frac{1}{b}$ are in $AP$,so $\frac{1}{a} + \frac{1}{b} = 2 	imes 10 = 20$.$\frac{a+b}{ab} = 20 \implies a+b = 20ab$ $(ii)$.Substitute $(i)$ into $(ii)$:$18b^2 + b = 20(18b^2)b = 360b^3$.Since $b > 0$,divide by $b$: $18b + 1 = 360b^2 \implies 360b^2 - 18b - 1 = 0$.Using the quadratic formula: $b = \frac{18 \pm \sqrt{324 - 4(360)(-1)}}{2(360)} = \frac{18 \pm \sqrt{324 + 1440}}{720} = \frac{18 \pm \sqrt{1764}}{720} = \frac{18 \pm 42}{720}$.Since $b > 0$,$b = \frac{60}{720} = \frac{1}{12}$.Then $a = 18 	imes (\frac{1}{12})^2 = 18 	imes \frac{1}{144} = \frac{1}{8}$.Finally,$16a + 12b = 16(\frac{1}{8}) + 12(\frac{1}{12}) = 2 + 1 = 3$.

For two positive numbers $a$ and $b$,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 $.........$.

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