Question

Find three numbers which form a geometric progression, if their product is $64$ and the arithmetic mean is $\frac{14}{3}$.

Answer 1

let th numbers be $\frac{a}{r},a,ar$

$⟹a^3=64⟹a=4$

$\frac{(\frac{a}{r}+a+ar)}{3}=14/3$

$4(r^2+r+1)=14r$

$2r^2-5r+2=0$

$⟹r=2,1/2$

$\therefore$ the numbers are $2,4,8$