If $3^a=27^b=81^c$ and $a b c=144$, then the value of $12\left(\frac{1}{a}+\frac{1}{2 b}+\frac{1}{5 c}\right)$ is: |
$\frac{18}{120}$ $\frac{18}{10}$ $\frac{33}{10}$ $\frac{17}{120}$ |
$\frac{33}{10}$ |
3a = 27b = 81c 3a = 33b = 34c If we put a = 12, b = 4 and c = 3, then 312 = 312 = 312 [satisfied] Now, $12\left(\frac{1}{a}+\frac{1}{2 b}+\frac{1}{5 c}\right)$ = $12\left(\frac{1}{12}+\frac{1}{8}+\frac{1}{15}\right)$ $12\left(\frac{1}{a}+\frac{1}{2 b}+\frac{1}{5 c}\right)$ = 12(\(\frac{10 + 15 + 8}{120}\)) $12\left(\frac{1}{a}+\frac{1}{2 b}+\frac{1}{5 c}\right)$ = 12(\(\frac{33}{120}\)) = $12\left(\frac{1}{a}+\frac{1}{2 b}+\frac{1}{5 c}\right)$ = $\frac{33}{10}$ |