i need 3 numbers that fill the conditions

3 numbers of continued proportion which the sum is 10 and the product of the first two is 6

2/3
27/3
1/3
?

Do you mean :

find x and k such that :

(x) + (kx) + (k²x) = 10
and
(x).(kx) = 6

?

my understanding would be (using a,b and c in sequence):

a+b+c=10, a*b=6 and b/a=c/b (this for the continued proportion)?

which means three equations with three unknown
go ahead and solve them

thanks thats what i wanted

Post back if you need help to solve this 4th degree equation.