# terms of a geometric series and arithmetic series, find common ratio

• May 21st 2012, 04:57 AM
itsgotabigbirdonitdude
terms of a geometric series and arithmetic series, find common ratio
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.
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.
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.
• May 21st 2012, 05:12 AM
Prove It
Re: terms of a geometric series and arithmetic series, find common ratio
Quote:

Originally Posted by itsgotabigbirdonitdude
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 fifth 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.
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 = 3(y-x) (since y-x is the common difference of the arithmetic progression and z-y is the difference between the second and fifth terms) and y-x = y/x.
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.

If \displaystyle \displaystyle \begin{align*} x, y, z \end{align*} are the first three terms of a geometric progression, then \displaystyle \displaystyle \begin{align*} x = t_1, y = r\,t_1, z &= r^2t_1 \end{align*}.

They are also the first, second and fifth terms of an arithmetic progression, so \displaystyle \displaystyle \begin{align*} x = t_1, y = t_1 + d \end{align*} and \displaystyle \displaystyle \begin{align*} z = t_1 + 4d \end{align*}.

Equating the values of \displaystyle \displaystyle \begin{align*} y \end{align*} gives

\displaystyle \displaystyle \begin{align*} r\,t_1 &= t_1 + d \\ r\,t_1 - t_1 &= d \end{align*}

and equating the values of \displaystyle \displaystyle \begin{align*} z \end{align*} gives

\displaystyle \displaystyle \begin{align*} r^2t_1 &= t_1 + 4d \\ r^2t_1 &= t_1 + 4\left(r\,t_1 - t_1\right) \\ r^2t_1 &= t_1 + 4r\,t_1 - 4t_1 \\ r^2t_1 &= 4r\,t_1 - 3t_1 \\ r^2t_1 - 4r\,t_1 + 3t_1 &= 0 \\ \left(r^2 - 4r + 3\right)t_1 &= 0 \\ t_1 = 0 \textrm{ or } r^2 - 4r + 3 &= 0 \\ (r - 1)(r - 3) &= 0 \\ r - 1 = 0 \textrm{ or } r - 3 &= 0 \\ r = 1 \textrm{ or } r &= 3 \end{align*}
• May 22nd 2012, 05:57 AM
itsgotabigbirdonitdude
Re: terms of a geometric series and arithmetic series, find common ratio
actually I had made a mistake in my first post, which I quickly edited but apparently Prove It didn't catch that in time. x, y and z are the first, second and FOURTH terms of an arithmetic series. Anyway 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?
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}