Question

For the two positive numbers $a, b$, if $a, b$ and $\frac{1}{18}$ are in a geometric progression, while $\frac{1}{a}, 10$ and $\frac{1}{b}$ are in an arithmetic progression, then $16 a+12 b$ is equal to

Answer 1

$ a , b , \frac{1}{18} \rightarrow GP$
$ \frac{ a }{18}= b ^2 ....$(i)
$ \frac{1}{ a }, 10, \frac{1}{ b } \rightarrow AP$
$ \frac{1}{ a }+\frac{1}{ b }=20$
$ \Rightarrow a + b =20 ab \text {, from eq. (i); we get }$
$\Rightarrow 18 b ^2+ b =360 b ^3$
$ \Rightarrow 360 b ^2-18 b -1=0 \{\because b \neq 0\} $
$ \Rightarrow b =\frac{18 \pm \sqrt{324+1440}}{720} $
$ \Rightarrow b =\frac{18+\sqrt{1764}}{720} \{\because b >0\} $
$\Rightarrow b =\frac{1}{12} $
$ \Rightarrow a =18 \times \frac{1}{144}=\frac{1}{8}$
Now, $16 a+12 b=16 \times \frac{1}{8}+12 \times \frac{1}{12}=3$