# Geometric and Arithmetic series question

• Apr 30th 2008, 07:43 PM
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 $\displaystyle 9 = x(r^3-1)/(r-1)$
A.S $\displaystyle 9 =(3/2) * (2x + (3-1)*d$

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...

• Apr 30th 2008, 08:26 PM
mr fantastic
Quote:

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

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...

"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.
• Apr 30th 2008, 08:39 PM
Soroban

Quote:

The numbers $\displaystyle x, y, z$ are succesive terms of a G.S.
with the sum of the three terms equal to 9.
When taken in the order $\displaystyle y, x ,z$, the numbers form an A.S.
Find the values of $\displaystyle x, y, z$

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

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

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

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

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

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

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

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

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

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

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