SOLUTION: Find 3 number in the geometrical progression whose sum is 28 and whose product is 512

Question 1066287: Find 3 number in the geometrical progression whose sum is 28 and whose product is 512
Found 2 solutions by swincher4391, Alan3354:
Answer by swincher4391(1107) (Show Source): You can put this solution on YOUR website!
Let a be the first number, b be the 2nd, c be the 3rd.
So a + b + c = 28 and a*b*c = 512.
Normally this would not be enough, but we know that there is a progression.
Let r be the common ratio between numbers in a geometric progression.
So the 2nd number can be written as:
b = r*a
And the 3rd number can be written as:
c = r^2*a
Go back to our original problem.
a + b + c can be written as a + ra + r^2*a
a*b*c can be written as a * (ra) * (r^2*a) = r^3*a^3
so we have that two equations.
a+ra+r^2*a = 28
r^3*a^3 = 512
We can solve for ra by taking the cube root of both sides in the 2nd equation.
(r*a)^3 = 512
(r*a) = 8
So a = 8/r
8/r + 8 + 8r = 28
8/r + 8r = 20
8(1/r + r) = 20
(r^2+1 / r) = 5/2
2(r^2+ 1) = 5r
2r^2 + 2 = 5r
2r^2 - 5r +2 = 0
we can rewrite as (2r-1)(r-2) so r = 1/2 or r = 2.
Let's try r = 2 first.
So if r =2 then a = 4
So we have
4, 8, 16 which do satisfy the conditions. But are there other numbers?
Try r = 1/2. Then a = 16.
So we have 16, 8 , 4. Ah, so it appears it is the same either way.
The numbers are 4, 8 , 16.

Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Find 3 number in the geometrical progression whose sum is 28 and whose product is 512
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Hint: 512 = 2^9
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