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.
Let , then . Since and are undefined . This means that we only need to find where the derivative equals , meaning . So , meaning that
For 2 positive real numbers and , . Let and . Then: . This implies that . or . Rearranging : . Thus, the smallest value is .
Solution 2
A geometric sequence always looks like
and they say that the second term . You should know that the sum of an infinite geometric series (denoted by here) is . We now have a system of equations which allows us to find in one variable.
We seek the smallest positive value of S. We proceed by graphing in the plane and find the answer is
We seek the smallest positive value of S. We proceed by graphing in the plane and find the answer is
We seek the smallest positive value of S. I'll be back soon to finish this solution. Please don't delete it. :(
We seek the smallest positive value of S. I'll be back soon to finish this solution. Please don't delete it. :(
See Also
2016 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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 12 Problems and Solutions |
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