Hello...
Find x and y if the sequence: 2y, 2xy, 2, xy/2 is geometric
Thank you for the help
Hello, IStandAlone99!
$\displaystyle \text{Find }x\text{ and }y\text{ if the sequence: }\,2y,\,2xy,\,2,\,\tfrac{xy}{2}\text{ is geometric.}$
We have: .$\displaystyle \begin{Bmatrix}a &=& 2y & [1] \\ ar &=& 2xy & [2] \\ ar^2 &=& 2 & [3] \\ ar^3 &=& \frac{xy}{2} & [4] \end{Bmatrix}$
$\displaystyle \begin{array}{ccccccccc}[2]\div[1]: & \dfrac{ar}{a} &=& \dfrac{2xy}{2y} & \Rightarrow & \;\;r &=& x & \; [5] \end{array}$
$\displaystyle \begin{array}{ccccccccc} [3]\div[2]: & \dfrac{ar^2}{ar} &=& \dfrac{2}{2xy} & \Rightarrow & r &=& \dfrac{1}{xy} & [6]\end{array}$
$\displaystyle \begin{array}{ccccccccc} [4]\div[3]: & \dfrac{ar^3}{ar^2} &=& \dfrac{\frac{xy}{2}}{2} & \Rightarrow & \;\;r &=& \dfrac{xy}{4} & [7] \end{array}$
$\displaystyle \begin{array}{ccccccccc}\text{Equate [7] and [6]:} & \frac{xy}{4} \,=\,\frac{1}{xy} & \Rightarrow & x^2y^2 \:=\:4 & [8] \\ \text{Equate [5] and [6]:} & x \,=\,\frac{1}{xy} & \Rightarrow & x^2y \:=\:1 & [9] \end{array}$
$\displaystyle [8] \div [9]: \;\;\frac{x^2y^2}{x^2y} \:=\:\frac{4}{1} \quad\Rightarrow\quad \boxed{y \,=\,4}$
Substitute into [9]: .$\displaystyle x^2(4) \,=\,1 \quad\Rightarrow\quad x^2 \,=\,\tfrac{1}{4} \quad\Rightarrow\quad \boxed{x \:=\:\pm\tfrac{1}{2}}$
Edit: Too slow again!
I believe my solution echoes parsum's ... just more details.