use the properties of this term tto find the fifteenth term of the geometric sequence 9x, x^2,2xy,8y...
+247 Using the properity of geometric equations that the ratio between terms is constant, you can set up the following eqation using the ratios between terms:
\(\frac{x^2}{9x}=\frac{2xy}{x^2} \Rightarrow x^{4}=18x^2y \Rightarrow x^{2}=18y \Rightarrow y=\frac{x^2}{18}\)
Creating another equation so you can solve gets this:
\(\frac{2xy}{x^2}=\frac{8y}{2xy} \Rightarrow 4x^2y^2=8x^2y \Rightarrow 4y^2=8y\Rightarrow y^2=4y\)
Remembering that an equation with a squared term has two solutions, we get y=4 and y=0
Plugging these values into the first equation yields x=0 and x=\(6\sqrt{2} \)
Thus, the first two terms of the sequence are \(54\sqrt{2}\) and 72 or 0 and 0
From here you should be able to find the common ratio R so the 15th term would be \(54\sqrt{2} \cdot R^{15-1}\) or 0
+26367 use the properties of this term tto find the fifteenth term of the geometric sequence
\(9x,\ x^2,\ 2xy,\ 8y,\ \ldots\)
I assume
\(\text{Let $r$ the common ratio }\\ \text{Let $a_1 = 9x$ } \\ \text{Let $a_2 = x^2 $ } \\ \text{Let $a_3 = 2xy $ } \\ \text{Let $a_4 = 8y $ } \\ \text{Let $a_{15} = \ ? $ } \)
\(\begin{array}{|rcll|} \hline r &=& \dfrac{a_2}{a_1} \\ &=& \dfrac{x^2}{9x} \\\\ \mathbf{r} &=& \mathbf{\dfrac{x}{9}} \qquad (1) \\ \mathbf{x^2} &=& \mathbf{9xr} \qquad (2) \\ \hline \end{array} \)
\(\begin{array}{|rcll|} \hline r &=& \dfrac{a_3}{a_2} \\ &=& \dfrac{2xy}{x^2} \\\\ \mathbf{2xy} &=& \mathbf{x^2r} \qquad (3) \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline r &=& \dfrac{a_4}{a_3} \\ &=& \dfrac{8y}{2xy} \\\\ \mathbf{2xy} &=& \mathbf{\dfrac{8y}{r} } \qquad (4) \\ \hline \end{array}\)
\(\mathbf{(3)=(4)}\)
\(\begin{array}{|rcll|} \hline 2xy = x^2r &=& \dfrac{8y}{r} \\ x^2r &=& \dfrac{8y}{r} \quad | \quad \mathbf{x^2=9xr} \qquad (2) \\ 9xr^2 &=& \dfrac{8y}{r} \\ \mathbf{r^3} &=& \mathbf{\dfrac{8y}{9x} } \qquad (5)\quad | \quad x\ne 0! \\ \hline \end{array} \)
\(\begin{array}{|rcll|} \hline r^3 = \left(\dfrac{x}{9}\right)^3 &=& \dfrac{8y}{9x} \quad | \quad \mathbf{r=\dfrac{x}{9}} \qquad (1), \qquad \mathbf{r^3=\dfrac{8y}{9x} } \qquad (5) \\ \left(\dfrac{x}{9}\right)^3 &=& \dfrac{8y}{9x} \\ x^4 &=& 9^2\cdot 8y \\ x^4 &=& 648y \\ \mathbf{y} &=& \mathbf{\dfrac{x^4}{648} } \qquad (6) \\ \hline \end{array}\)
\(\text{Let $a_1 = a = 9x$ } \\ \text{Let $a_2 = ar = 9xr $ } \\ \text{Let $a_3 = ar^2 = 9xr^2 $ } \\ \text{Let $a_4 = ar^3 $ } \\ \text{Let $a_{15} = ar^{14}=9xr^{14} $ } \)
\(\mathbf{x=\ ? }\)
\(\begin{array}{|rcll|} \hline a_3 &=& 9xr^2 \quad | \quad a_3 = 2xy \\ 2xy &=& 9xr^2 \quad | \quad \mathbf{r=\dfrac{x}{9}} \qquad (1) \\ 2xy &=& 9x\left(\dfrac{x}{9}\right)^2 \\ 2xy &=& \dfrac{x^3}{9} \\ x^3 &=& 18xy \quad | \quad \mathbf{y=\dfrac{x^4}{648} } \qquad (6) \\ x^3 &=& 18x\dfrac{x^4}{648} \\ x^3 &=& \dfrac{18x^5}{648} \\ x^2 &=& \dfrac{648}{18} \\ \mathbf{x^2} &=& \mathbf{36} \qquad \mathbf{x=\pm6} \\ \hline \end{array}\)
\(\mathbf{y=\ ? }\)
\(\begin{array}{|rcll|} \hline \mathbf{y} &=& \mathbf{\dfrac{x^4}{648} } \quad | \quad \mathbf{x^2=36} \\ y &=& \dfrac{36^2}{648} \\ \mathbf{y} &=& \mathbf{2} \\ \hline \end{array}\)
1. geometric sequence
\(x=6,\ y=2,\ r=\dfrac{x}{9}=\dfrac{6}{9}=\dfrac{2}{3} \)
\(\begin{array}{|rcll|} \hline a_1 &=& 54 \\ a_2 &=& 36 \\ a_3 &=& 24 \\ a_4 &=& 16 \\\\ a_{15} &=& 54\cdot \left( \dfrac{2}{3} \right)^{14} \\\\ &=& \dfrac{54}{3^3}\cdot \dfrac{2^{14}}{3^{11}} \\\\ &=& 2\cdot \dfrac{2^{14}}{3^{11}} \\\\ &=& \dfrac{2^{15}}{3^{11}} \\\\ &=& \dfrac{32768}{177147} \\\\ \mathbf{a_{15}} &=& \mathbf{ 0.18497631910 } \\ \hline \end{array}\)
2. geometric sequence
\(x=-6,\ y=2,\ r=\dfrac{x}{9}=\dfrac{-6}{9}=-\dfrac{2}{3}\)
\(\begin{array}{|rcll|} \hline a_1 &=& -54 \\ a_2 &=& 36 \\ a_3 &=& -24 \\ a_4 &=& 16 \\\\ a_{15} &=& -54\cdot \left( -\dfrac{2}{3} \right)^{14} \\\\ &=& -\dfrac{54}{3^3}\cdot \dfrac{2^{14}}{3^{11}} \\\\ &=& -2\cdot \dfrac{2^{14}}{3^{11}} \\\\ &=& - \dfrac{2^{15}}{3^{11}} \\\\ &=& -\dfrac{32768}{177147} \\\\ \mathbf{a_{15}} &=& \mathbf{ -0.18497631910 } \\ \hline \end{array}\)
