Alg8.13

Three numbers are in G.P.Their sum is $21\,$ and their product is $216\,$ .Find them.

Let the three numbers be $\frac{a}{r},a,ar\,$

Their product is $a^3=216,a=6\,$

Their sum is $\frac{6}{r}+6+6r=21\,$

$6r^2+6r+6=21r\,$

$6r^2-15r+6=0,2r^2-5r+2=0,(2r-1)(r-2)=0,r=\frac{1}{2},2\,$

If $r=\frac{1}{2}\,$ the numbers are $12,6,3\,$

If $r=2\,$ the numbers are $3,6,12\,$