2008 AMC 12A Problems/Problem 16
Contents
Problem
The numbers , , and are the first three terms of an arithmetic sequence, and the term of the sequence is . What is ?
Solutions
Solution 1
Let and .
The first three terms of the arithmetic sequence are , , and , and the term is .
Thus, .
Since the first three terms in the sequence are , , and , the th term is .
Thus the term is .
Solution 2
If , , and are in arithmetic progression, then , , and are in geometric progression. Therefore,
Therefore, , , therefore the 12th term in the sequence is
Solution 3
Solution 4 (mimimal manipulation)
Given the first three terms form an arithmetic progression, we have: Subtracting the first equation from the second and the third from the second, respectively, gives us these two expressions for : The desired th term in the sequence is , so we can substitute our values for and (using either one of our two expressions for ): The answer must be expressed as , however. We're in luck: the two different yet equal expressions for allow us to express and in terms of each other: Plugging in , we have: ~ Jingwei325
See Also
2008 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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All AMC 12 Problems and Solutions |
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