Question

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

Answer 1

Answer: 3

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