Difference between revisions of "2003 AMC 10B Problems/Problem 24"
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<math> \textbf{(A)}\ -\frac{15}{8}\qquad\textbf{(B)}\ -\frac{6}{5}\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ \frac{27}{20}\qquad\textbf{(E)}\ \frac{123}{40} </math> | <math> \textbf{(A)}\ -\frac{15}{8}\qquad\textbf{(B)}\ -\frac{6}{5}\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ \frac{27}{20}\qquad\textbf{(E)}\ \frac{123}{40} </math> | ||
| − | =Solution = | + | ==Solution== |
The difference between consecutive terms is <math>(x-y)-(x+y)=-2y.</math> Therefore we can also express the third and fourth terms as <math>x-3y</math> and <math>x-5y.</math> Then we can set them equal to <math>xy</math> and <math>\frac{x}{y}</math> because they are the same thing. | The difference between consecutive terms is <math>(x-y)-(x+y)=-2y.</math> Therefore we can also express the third and fourth terms as <math>x-3y</math> and <math>x-5y.</math> Then we can set them equal to <math>xy</math> and <math>\frac{x}{y}</math> because they are the same thing. | ||
Revision as of 17:54, 21 October 2020
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.
![\[\frac{x}{y}=x-5y\]](http://latex.artofproblemsolving.com/7/4/4/7447ff4042273b628b683c0be7a6adb782094cc4.png)
![\[\frac{-3}{y-1}=\frac{-3y}{y-1}-5y\]](http://latex.artofproblemsolving.com/e/a/f/eafd224e1b0406671c0ea78a89f70250ab4d9598.png)
![\[-3=-3y-5y(y-1)\]](http://latex.artofproblemsolving.com/5/8/2/582e642d7ac3fe394341c104bb678404cf1937d6.png)
![\[0=5y^2-2y-3\]](http://latex.artofproblemsolving.com/4/f/c/4fc65b027cacf27699496e8f7a46167eb301ea72.png)
![\[0=(5y+3)(y-1)\]](http://latex.artofproblemsolving.com/2/7/1/2715b828ebf0e900ba0d1a4a4409393b2d5c9683.png)
![\[y=-\frac35, 1\]](http://latex.artofproblemsolving.com/2/b/f/2bf85034e5c754150e8e99435ec811627279e560.png)
But 
cannot be 
because then the first term would be 
and the second term 
while the last two terms 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
