# Math Help - Find x and y if the sequence is geometric

1. ## Find x and y if the sequence is geometric

Hello...
Find x and y if the sequence: 2y, 2xy, 2, xy/2 is geometric

Thank you for the help

2. ## Re: Find x and y if the sequence is geometric

if series is geometric
there is a common ratio r such that
a,ar,ar^2,... represent geometric series
now here r=2xy/2y=2/2xy
also 2/2xy=(xy/2)/2
solving both these u get
x=1/2 and y=4

3. ## Re: Find x and y if the sequence is geometric

Originally Posted by IStandAlone99
Hello...
Find x and y if the sequence: 2y, 2xy, 2, xy/2 is geometric

Thank you for the help
Let's take this in pieces. What is the general formula for a geometric series? How many variables does that general formula have?

prasum beat me to it.

4. ## Re: Find x and y if the sequence is geometric

Hello, IStandAlone99!

$\text{Find }x\text{ and }y\text{ if the sequence: }\,2y,\,2xy,\,2,\,\tfrac{xy}{2}\text{ is geometric.}$

We have: . $\begin{Bmatrix}a &=& 2y & [1] \\ ar &=& 2xy & [2] \\ ar^2 &=& 2 & [3] \\ ar^3 &=& \frac{xy}{2} & [4] \end{Bmatrix}$

$\begin{array}{ccccccccc}[2]\div[1]: & \dfrac{ar}{a} &=& \dfrac{2xy}{2y} & \Rightarrow & \;\;r &=& x & \; [5] \end{array}$
$\begin{array}{ccccccccc} [3]\div[2]: & \dfrac{ar^2}{ar} &=& \dfrac{2}{2xy} & \Rightarrow & r &=& \dfrac{1}{xy} & [6]\end{array}$
$\begin{array}{ccccccccc} [4]\div[3]: & \dfrac{ar^3}{ar^2} &=& \dfrac{\frac{xy}{2}}{2} & \Rightarrow & \;\;r &=& \dfrac{xy}{4} & [7] \end{array}$

$\begin{array}{ccccccccc}\text{Equate [7] and [6]:} & \frac{xy}{4} \,=\,\frac{1}{xy} & \Rightarrow & x^2y^2 \:=\:4 & [8] \\ \text{Equate [5] and [6]:} & x \,=\,\frac{1}{xy} & \Rightarrow & x^2y \:=\:1 & [9] \end{array}$

$[8] \div [9]: \;\;\frac{x^2y^2}{x^2y} \:=\:\frac{4}{1} \quad\Rightarrow\quad \boxed{y \,=\,4}$

Substitute into [9]: . $x^2(4) \,=\,1 \quad\Rightarrow\quad x^2 \,=\,\tfrac{1}{4} \quad\Rightarrow\quad \boxed{x \:=\:\pm\tfrac{1}{2}}$

Edit: Too slow again!
I believe my solution echoes parsum's ... just more details.