Let 1/16, a and b be in G.P. and


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________.


$a^{2}=\frac{b}{16} \Rightarrow \frac{1}{b}=\frac{1}{16 a^{2}}$


$\frac{1}{8 \mathrm{a}^{2}}=\frac{1}{\mathrm{a}}+6$


$\frac{1}{\mathrm{a}}=12,-4 \Rightarrow \mathrm{a}=\frac{1}{12},-\frac{1}{4}$

$\frac{1}{a}=12,-4 \Rightarrow a=\frac{1}{12},-\frac{1}{4}$

$a=\frac{1}{12}, a>0$

$\mathrm{b}=16 \mathrm{a}^{2}=\frac{1}{9}$

$\Rightarrow 72(a+b)=6+8=14$


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