Question

Let $\frac{1}{16}$, $a$ and $b$ be in G.P. and $\frac{1}{a},\frac{1}{b},6$ be in A.P., where $a,b>0$. Then $72(a+b)$ is equal to ______.

Answer 1

Answer: 14

$a^2=\frac{b}{16}\Rightarrow \frac{1}{b}=\frac{1}{16a^2}$
$\frac{2}{b}=\frac{1}{a}+6$
$\frac{1}{8a^2}=\frac{1}{a}+6$
$\frac{1}{a^2}-\frac{8}{a}-48=0$
$\frac{1}{a}=12,-4\Rightarrow a=\frac{1}{12},-\frac{1}{4}$
$a=\frac{1}{12},a>0$
$b=16a^2=\frac{1}{9}$
$\Rightarrow 72(a+b)=6+8=14$