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:
$a^{2}=\frac{b}{16} \Rightarrow \frac{1}{b}=\frac{1}{16 a^{2}}$
$\frac{2}{b}=\frac{1}{a}+6$
$\frac{1}{8 a^{2}}=\frac{1}{a}+6$
$\frac{1}{a^{2}}-\frac{8}{a}-48=0$
$\frac{1}{a}=12,-4 \Rightarrow a=\frac{1}{12},-\frac{1}{4}$
$a=\frac{1}{12}, a>0$
$b=16 a^{2}=\frac{1}{9}$
$\Rightarrow \quad 72(a+b)=6+8=14$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.