# Prove the following

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:

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

$\frac{2}{b}=\frac{1}{a}+6$

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

$\frac{1}{a^{2}}-\frac{8}{a}-48=0$

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

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

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

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