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April 30th, 2008, 09:43 PM
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| |  Geometric and Arithmetic series question
The question is:
The numbers x, y, z are succesive terms of a G.S with the sum of the three terms equal to 9. When taken in the order y, x ,z the numbers form an A.S. Find the values of x, y, z?
G.S 
A.S 
So I have that the sum of the G.S and A.S are equal and therefore I can say that the equations are equal to each other but i just don't know where to go from here or if I'm correct...
Thanks for any help you can give me |
April 30th, 2008, 10:26 PM
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Quote:
Originally Posted by smplease  The question is:
The numbers x, y, z are succesive terms of a G.S with the sum of the three terms equal to 9. When taken in the order y, x ,z the numbers form an A.S. Find the values of x, y, z?
G.S
A.S
So I have that the sum of the G.S and A.S are equal and therefore I can say that the equations are equal to each other but i just don't know where to go from here or if I'm correct...
Thanks for any help you can give me | "The numbers x, y, z are succesive terms of a G.S ...." => y/x = z/y => y^2 = xz .... (1)
".... with the sum of the three terms equal to 9" => x + y + z = 9 .... (2)
"When taken in the order y, x ,z the numbers form an A.S." => x - y = z - x => 2x = y + z .... (3)
Solve equations (1), (2) and (3) simultaneously for x, y, z.
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April 30th, 2008, 10:39 PM
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Hello, smplease! The three terms are: .
Their sum is 9: .![a + ar + ar^2\:=\:9\quad\Rightarrow\quad a(1 + r + r^2) \:=\:9\;\;{\color{blue}[1]}](http://www.mathhelpforum.com/math-help/latex2/img/3579db9a1efb29ca3273ce500e92e29f-1.gif "a + ar + ar^2\:=\:9\quad\Rightarrow\quad a(1 + r + r^2) \:=\:9\;\;{\color{blue}[1]}")
We are told: . form an A.S.
If the common difference is  , we have: .![\begin{array}{cccccccc}ar + d &=& a & \Rightarrow & d &=&a(1-r) & {\color{blue}[2]}\\ a +d &=& ar^2 & \Rightarrow & d &=& a(r^2-1) & {\color{blue}[3]}\end{array}](http://www.mathhelpforum.com/math-help/latex2/img/04b8e972346d793baf5cdf320a5bf973-1.gif "\begin{array}{cccccccc}ar + d &=& a & \Rightarrow & d &=&a(1-r) & {\color{blue}[2]}\\ a +d &=& ar^2 & \Rightarrow & d &=& a(r^2-1) & {\color{blue}[3]}\end{array}")
Eqsuate [2] and [3]: .
. . 
If  , we have the trivial sequence: .
If  , substitute into [1]: .
The geometric sequence is: . . . . which has a sum of 12.
The arithmetic sequence is: . . . . which has 
Therefore: . 
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