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The first four terms in an arithmetic sequence are x + y, x - y, xy, and x/y, in that order. What is the fifth term? Express your answer as a common fraction.

 Mar 1, 2020
 #1
avatar+25226 
+2

The first four terms in an arithmetic sequence are x + y, x - y, xy, and x/y, in that order.
What is the fifth term?
Express your answer as a common fraction.

 

\(\begin{array}{rclcl} a_1 &=& a &=& x+y \\ a_2 &=& a+d &=& x-y \\ a_3 &=& a+2d &=& xy \\ a_4 &=& a+3d &=& \dfrac{x}{y} \quad &| \quad y\neq 0 ~!\\ a_5 &=& a+4d &=& \ ? \\ \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{a_2} &=& \mathbf{a+d} \quad &| \quad a=x+y\\ a_2 &=& x+y+d \\ a_2-(x+y) &=&d \quad &| \quad a_2 = x-y \\ x-y-(x+y) &=&d \\ \mathbf{-2y} &=&\mathbf{d}\qquad (1) \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{a_3} &=& \mathbf{a+2d} \quad &| \quad a=x+y \\ a_3 &=& x+y+2d \\ a_3-(x+y) &=& 2d \quad &| \quad a_3 = xy \\ \mathbf{ xy-(x+y)} &=& \mathbf{2d}\qquad (2) \\ \hline \end{array} \)

 

\(\begin{array}{|lrcll|} \hline \dfrac{(2)}{(1)}: & \mathbf{ \dfrac{xy-(x+y)}{-2y}} &=& \mathbf{ \dfrac{2d}{d}} \\\\ & \dfrac{xy-(x+y)}{-2y} &=& 2 \\ & xy-(x+y) &=& -4y \\ & xy-x-y &=& -4y \\ & xy-x+3y &=& 0 \\ &\mathbf{ x-xy } &=& \mathbf{3y} \qquad (3) \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{a_4} &=& \mathbf{a+3d} \quad &| \quad a=x+y \\ a_4 &=& x+y+3d \\ a_4-(x+y) &=& 3d \quad &| \quad a_4 = \dfrac{x}{y} \\ \mathbf{\dfrac{x}{y}-(x+y)} &=& \mathbf{3d} \qquad (4) \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline \dfrac{(4)}{(1)}: & \mathbf{ \dfrac{\dfrac{x}{y}-(x+y)}{-2y}} &=& \mathbf{ \dfrac{3d}{d}} \\\\ & \dfrac{\dfrac{x}{y}-(x+y)}{-2y} &=& 3 \\ & \dfrac{x}{y}-(x+y) &=& -6y \quad &| \quad *y \\ & x-(x+y)y &=& -6y^2 \\ & x-xy-y^2 &=& -6y^2 \\ & x-xy+5y^2 &=& 0 \quad &| \quad \mathbf{ x-xy =3y} \qquad (3) \\ & 3y+5y^2 &=& 0 \\ & y(3+5y) &=& 0 \quad &| \quad y\neq 0 ~!\\ & 3+5y &=& 0 \\ & 5y &=& -3 \\ & \mathbf{y} &=& \mathbf{-\dfrac{3}{5}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{d} &=& \mathbf{-2y} \qquad (1) \\ d &=& -2\left(-\dfrac{3}{5}\right) \\ \mathbf{d} &=& \mathbf{ \dfrac{6}{5} } \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{ xy-(x+y)} &=& \mathbf{2d}\qquad (2) \\ \ldots \\ x &=& \dfrac{2d+y}{y-1} \quad & | \quad \mathbf{y=-\dfrac{3}{5}},\ \mathbf{d= \dfrac{6}{5} } \\ x &=& \dfrac{2 *\dfrac{6}{5}-\dfrac{3}{5}}{-\dfrac{3}{5}-1} \\ x &=& -\dfrac{9}{5}* \dfrac{5}{8} \\ \mathbf{x} &=& \mathbf{-\dfrac{9}{8}} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline a &=& x+y \\ a &=& -\dfrac{9}{8}-\dfrac{3}{5} \\ \mathbf{a} &=& \mathbf{-\dfrac{69}{40}} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline a_5 &=& a+4d \\ a_5 &=& -\dfrac{69}{40} + 4*\dfrac{6}{5} \\ a_5 &=& -\dfrac{69}{40} + \dfrac{24}{5} \\ a_5 &=& -\dfrac{69}{40} + \dfrac{24*8}{5*8} \\ \mathbf{a_5} &=& \mathbf{\dfrac{123}{40}} \\ \hline \end{array}\)

 

arithmetric sequence: \(\{-\dfrac{69}{40},~-\dfrac{21}{40},~\dfrac{27}{40},~\dfrac{75}{40},~\dfrac{123}{40},~\ldots\}\)

 

laugh

 Mar 2, 2020
 #2
avatar+108 
+1

Thank you very much

brianlaw  Mar 3, 2020

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