2003 AMC 10B Problems/Problem 24
Contents
Problem
The first four terms in an arithmetic sequence are , , , and , in that order. What is the fifth term?
Solution 1
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 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,
Solution 2
Because this is an arithmetic sequence, we conclude from the first two terms that the common difference is . Therefore, and . If we multiply and , we see:
Because , by basic multiplication, is , we have
Now that we have in terms of , we substitute in in for in (the fourth term). This leaves us with .
Recall that can be written as . Thus, . Substitute in in for , and we see:
Aha! This means , the common difference, is . Now, all we need to do is find the fifth term, which is just . We can substitute known values to solve:
.
~SXWang
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.