Answer
Insert : 10, 20, 40
Work Step by Step
Let $a_{1},a_{2},a_{3},a_{4},a_{5 }$ be a geometric sequence,
for which we know $a_{1}=5$ and $a_{5}=80.$
We want to find $a_{2},a_{3},a_{4}$ such that all three lie between 5 and 80.
General term of a geometric sequence: $a_{n}=a_{1}r^{n-1}$,
so from $a_{5}=80$ we find r
$80=5r^{4}\quad/\div 5$
$r^{4}=16$
$r=2$ or $r=-2$
If $r=-2$,
$a_{2}=5\times(-2)=-10$,
...but this is not between 5 and 80 (not a geometric mean)
So, $r\neq-2$
Therefore, r $=2$, and
$a_{2}=5(2)=10$,
$a_{3}=10(2)=20$,
$a_{4}=20(2)=40$.
(all are between 5 and 80, so they are three geometric means)
