Difference between revisions of "2003 AMC 10B Problems/Problem 24"
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<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 | + | 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 values for <math>y</math> into any of the equations, we get <math>x=-\frac98.</math> Finally, |
<cmath> \frac{x}{y}-2y=\frac{9\cdot 5}{8\cdot 3}+\frac{6}{5}=\boxed{\textbf{(E)}\ \frac{123}{40}}</cmath> | <cmath> \frac{x}{y}-2y=\frac{9\cdot 5}{8\cdot 3}+\frac{6}{5}=\boxed{\textbf{(E)}\ \frac{123}{40}}</cmath> |
Revision as of 08:30, 2 August 2019
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 values 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.