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

$a^2=\frac{b}{16}$  and $\frac{2}{b}=\frac{1}{a}+6$ Solving, we get $a=\frac{1}{12}$ or $a=\frac{-1}{4}$ [rejected]
If  $a=\frac{1}{12}\Rightarrow b=\frac{1}{9}$   $\therefore 72(a+b) =72(\frac{1}{12}+\frac{1}{9}) =14$