Geometric and Arithmetic series question

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April 30th 2008, 07:43 PM
smplease

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 $9 = x(r^3-1)/(r-1)$
A.S $9 =(3/2) * (2x + (3-1)*d$

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, 08:26 PM
mr fantastic

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 $9 = x(r^3-1)/(r-1)$
A.S $9 =(3/2) * (2x + (3-1)*d$

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.
April 30th 2008, 08:39 PM
Soroban

Hello, smplease!

Quote:

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 $x, y, z$

The three terms are: . $a,\:ar,\:ar^2$

Their sum is 9: . $a + ar + ar^2\:=\:9\quad\Rightarrow\quad a(1 + r + r^2) \:=\:9\;\;{\color{blue}[1]}$

We are told: . $ar,\:a,\:ar^2$ form an A.S.

If the common difference is $d$ , we have: . $\begin{array}{cccccccc}ar + d &=& a & \Rightarrow & d &=&a(1-r) & {\color{blue}[2]}\\ a +d &=& ar^2 & \Rightarrow & d &=& a(r^2-1) & {\color{blue}[3]}\end{array}$

Eqsuate [2] and [3]: . $a(1-r) \:=\:a(r^2-1)\quad\Rightarrow\quad r^2 + r - 2 \:=\:0$

. . $(r-1)(r+2) \:=\:0\quad\Rightarrow\quad r \:=\:1,\:-2$

If $r = 1$ , we have the trivial sequence: . $3,\:3,\:3$

If $r = -2$ , substitute into [1]: . $a(1 - 2 + 4) \:=\:9\quad\Rightarrow\quad a \:=\:3$

The geometric sequence is: . $3,\:-6,\:12$ . . . which has a sum of 12.

The arithmetic sequence is: . $-6,\:3,\:12$ . . . which has $d = 9$

Therefore: . $\begin{Bmatrix}x &=& 3 \\ y &=&\text{-}6 \\ z &=& 12 \end{Bmatrix}$