# Continuous ratio

1. Jul 10, 2016

### Joseph Richard

1. The problem statement, all variables and given/known data
Determine the conditions of a continuous ratio knowing that the product of the four terms is 1296 and the last term is equal to 1/6 of the sum of means.

Original question (in Portuguese):
Determinar as condições de uma proporção contínua sabendo que o produto dos quatro termos é 1296 e o último termo é igual a 1/6 da soma dos meios.

2. The attempt at a solution
x/y=y/w = 1296
w = 2y/6 , x = 1296y, and then, I tried to develop, and it gave a mess...
I would like that you guys help me.
Thank you very much in advance

Template: 18, 6, 6 and 2.

Last edited: Jul 10, 2016
2. Jul 10, 2016

### SammyS

Staff Emeritus
Hello Joseph Richard, Welcome to PF !

What is the definition of a "continuous ratio"? I'm not familiar with that terminology.

3. Jul 10, 2016

### Joseph Richard

Hi Sammy,
I don't know if I translated wrong, but this is continuous ratio is one in which the means or the ends are the same, the party ratio and proportion
Example:
9/6 = 6/4

4. Jul 10, 2016

### haruspex

I would guess we have four consecutive terms in geometric series. But I'm baffled by "sum of the means". Is this the sum of the pairwise geometric means? Of the pairwise arithmetic means? Of the middle two terms? ...?

5. Jul 10, 2016

### Joseph Richard

Harus, I don't know too.

6. Jul 10, 2016

### Joseph Richard

I know, in the subject of Arithmetic, in the part of ratio and proportion, sum of the means has to do about the means and extremes.

7. Jul 10, 2016

### Joseph Richard

x/y=y/w
y²=xw
(y²)²=1296
y=6

1/6 of y+y ----> 1/6 de 12 => 2 = w
Replacing is
36=2x
x = 18
S = {6, 6, 18, 2}

8. Jul 10, 2016

### Joseph Richard

Now that I solved the question, I was on it for three days.
Thank you to everyone who helped me.

9. Jul 10, 2016

### haruspex

Should that be (18, 6, 6, 2)?

10. Jul 12, 2016

### Joseph Richard

Yes Haurs, I messed up.
Thank you for the correction.