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

thekopite
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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?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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