Terms of a geometric series and arithmetic series, find common ratio

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Homework Help Overview

The problem involves determining the common ratio of a geometric progression where the first three terms are represented by different variables, and these terms also serve as specific terms in an arithmetic progression. The original poster is struggling primarily with the first part of the problem, which requires finding the common ratio.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to establish relationships between the terms of the geometric and arithmetic progressions using equations derived from their definitions. Some participants suggest manipulating these equations to derive a second-order equation for the common ratio.

Discussion Status

Participants are actively engaging with the problem, with some providing suggestions for how to approach the equations. The original poster has made progress on part a, identifying potential values for the common ratio, but is still seeking guidance on part b, indicating a productive exchange of ideas without a clear consensus on the next steps.

Contextual Notes

The original poster notes confusion regarding the combination of equations and the need to identify specific terms in the arithmetic progression, raising questions about the necessity of finding additional variables.

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Homework Statement



Different numbers x, y and z are the first three terms of a geometric progression with common ratio r, and also the first, second and fourth terms of an arithmetic progression.
a. Find the value of r.
b. Find which term of the arithmetic progression will next be equal to a term of the geometric progression.

I haven't tackled part b. yet but I'm guessing it must be quite straightforward once r is found, but for now I'm having major issues with a.

Homework Equations



So far I've come up with r=y/x=z/y (since all are different versions of the common ratio of the geometric progression), z-y = 2(y-x) (since y-x is the common difference of the arithmetic progression and z-y is the difference between the second and fourth terms) and y-x = y/x.

The Attempt at a Solution



However, I'm confused as to how to combine these equations in order to find r. All of my attempts have turned up hopelessly complex or just plain incorrect. Any suggestions as to the correct and most simple way to go about this would be appreciated, cheers.
 
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What you are doing looks entirely reasonable. So suppose you write
r = z/y, and then solve z = 3y-2x and plug it back in. You get a 2nd order equation for r.
 
So I managed to solve a. in the following manner:
r=y/x which means that y=rx, and since r=z/y, z=ry=(r^2)x
also z-y = 2(y-x), and plugging in the above values (in terms of x) of z and y I get x(r^2) - rx = 2rx - 2x
eliminating x from both sides of this equation gives (r^2) - 3r + 2 = 0, which gives me r=1 or 2.

However I'm having some problems with b.
I'm using the equation a(r^(n-1)) = a + (n-1) d, where a=x, r=2 and I need to solve for n.
Other than that, so far I've just been working in circles, and it's pretty frustrating. Where should I start looking for x and d? Do I even need to find x?
 
What's the corresponding increase you make to the arithmetic sequence for each value of r? So if r=2, what does the arithmetic sequence look like?
 

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