Difference between revisions of "2003 AMC 10B Problems/Problem 24"
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<cmath> | <cmath> | ||
− | \frac{x}{y}=x-5y | + | \frac{x}{y}=x-5y</cmath> |
− | \frac{-3}{y-1}=\frac{-3y}{y-1}-5y | + | <cmath>\frac{-3}{y-1}=\frac{-3y}{y-1}-5y</cmath> |
− | -3=-3y-5y(y-1) | + | <cmath>-3=-3y-5y(y-1)</cmath> |
− | 0=5y^2-2y-3 | + | <cmath>0=5y^2-2y-3</cmath> |
− | 0=(5y+3)(y-1) | + | <cmath>0=(5y+3)(y-1)</cmath> |
− | y=-\frac35, 1</cmath> | + | <cmath>y=-\frac35, 1</cmath> |
But <math>y</math> cannot be <math>1</math> because then every term would be equal to <math>x.</math> Therefore <math>y=-\frac35.</math> Substituting the value for <math>y</math> into any of the equations, we get <math>x=-\frac98.</math> Finally, | But <math>y</math> cannot be <math>1</math> because then every term would be equal to <math>x.</math> Therefore <math>y=-\frac35.</math> Substituting the value for <math>y</math> into any of the equations, we get <math>x=-\frac98.</math> Finally, |
Revision as of 00:16, 30 December 2015
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.
But cannot be because then every term would be equal to Therefore Substituting the value for into any of the equations, we get Finally,
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.