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

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Homework Help Overview

The problem involves determining the conditions of a continuous ratio with the constraints that the product of four terms equals 1296 and that the last term is one-sixth of the sum of the means. The subject area pertains to ratios and proportions, specifically within the context of continuous ratios.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the definition of a "continuous ratio" and its implications, with some suggesting it may relate to geometric series. Questions arise regarding the interpretation of "sum of the means" and whether it refers to arithmetic or geometric means.

Discussion Status

The discussion includes attempts to derive relationships between the terms based on the given conditions. Some participants express confusion about terminology and definitions, while others provide insights into potential approaches. A few participants have made calculations leading to specific values for the terms, but there is no consensus on the interpretation of the problem.

Contextual Notes

There is mention of a template for the terms, and some participants express uncertainty about the definitions and relationships involved in the problem. The original poster requests help after encountering difficulties in their attempts to solve the problem.

Joseph Richard
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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.