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