Difference between revisions of "2003 AMC 10B Problems/Problem 24"
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\frac{-3}{y-1}&=\frac{-3y}{y-1}-5y\\ | \frac{-3}{y-1}&=\frac{-3y}{y-1}-5y\\ | ||
-3&=-3y-5y(y-1)\\ | -3&=-3y-5y(y-1)\\ | ||
| − | 0&=5y^2-2y- | + | 0&=5y^2-2y-3\\ |
0&=(5y+3)(y-1)\\ | 0&=(5y+3)(y-1)\\ | ||
y&=-\frac35, 1</cmath> | y&=-\frac35, 1</cmath> | ||
Revision as of 22:30, 25 June 2013
Problem
The first four terms in an arithmetic sequence are 
,
,
, and 
, in that order. What is the fifth term?
Solution
The difference between consecutive terms is 
Therefore we can also express the third and fourth terms as 
and 
Then we can set them equal to and because they are the same thing.
Substitute into our other equation.
\begin{align*}
\frac{x}{y}&=x-5y\\
\frac{-3}{y-1}&=\frac{-3y}{y-1}-5y\\
-3&=-3y-5y(y-1)\\
0&=5y^2-2y-3\\
0&=(5y+3)(y-1)\\
y&=-\frac35, 1 (Error compiling LaTeX. Unknown error_msg)
But 
cannot be 
because then every term would be equal to 
Therefore 
Substituting the value for into any of the equations, we get 
Finally,
![\[\frac{x}{y}-2y=\frac{9\cdot 5}{8\cdot 3}+\frac{6}{5}=\boxed{\textbf{(E)}\ \frac{123}{40}}\]](http://latex.artofproblemsolving.com/5/5/e/55e30872e72c0aa052d66c068d2af833a675a508.png)
See Also
| 2003 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
