2016 AMC 12B Problems/Problem 14
Contents
Problem
The sum of an infinite geometric series is a positive number , and the second term in the series is . What is the smallest possible value of
Solution
The second term in a geometric series is , where is the common ratio for the series and is the first term of the series. So we know that and we wish to find the minimum value of the infinite sum of the series. We know that: and substituting in , we get that . From here, you can either use calculus or AM-GM.
Calculus: Let , then . Since and are undefined . This means that we only need to find where the derivative equals , meaning . So , meaning that
AM-GM For 2 positive real numbers and , . Let and . Then: . This implies that . or . Rearranging : . Thus, the smallest value is .
Solution 2
A simple approach is to initially recognize that and . We know that , since the series must converge. We can start by observing the greatest answer choice, 4. Therefore, , because that would make , which would make the series exceed 4. In order to minimize both the initial term and the rest of the series, we can recognize that is the opitimal ratio, thus the answer is .
See Also
2016 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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All AMC 12 Problems and Solutions |
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