Continuous Ratio Conditions: Product of 1296, Last Term 1/6 of Sum of Means

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The discussion revolves around determining the conditions of a continuous ratio where the product of four terms is 1296, and the last term equals 1/6 of the sum of the means. Participants clarify the definition of a continuous ratio, suggesting it involves four consecutive terms in a geometric series. Through calculations, one user finds the terms to be 18, 6, 6, and 2, confirming that these satisfy the given conditions. The conversation highlights the confusion surrounding the terminology and the mathematical approach to solving the problem. Ultimately, the solution was reached after collaborative effort and clarification.
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Homework Statement


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:
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Joseph Richard said:

Homework Statement


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 das 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.
Hello Joseph Richard, Welcome to PF !

What is the definition of a "continuous ratio"? I'm not familiar with that terminology.
 
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
 
SammyS said:
Hello Joseph Richard, Welcome to PF !

What is the definition of a "continuous ratio"? I'm not familiar with that terminology.
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? ...?
 
Harus, I don't know too. :cry:
 
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.
 
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}
 
Now that I solved the question, I was on it for three days.
Thank you to everyone who helped me.
 
Joseph Richard said:
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}
Should that be (18, 6, 6, 2)?
 
  • #10
Yes Haurs, I messed up.
Thank you for the correction.
 

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