2016 AMC 10B Problems/Problem 16
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 sum of an infinite geometric series is of the form: where is the first term and is the ratio whose absolute value is less than 1.
We know that the second term is the first term multiplied by the ratio. In other words:
Thus, the sum is the following:
Since we want the minimum value of this expression, we want the maximum value for the denominator, . The maximum x-value of a quadratic with negative is .
Plugging into the quadratic yields:
Therefore, the minimum sum of our infinite geometric sequence is .
Solution 2
After observation we realize that in order to minimize our sum with being the reciprocal of r, the common ratio has to be in the form of with being an integer as anything more than divided by would give a larger sum than a ratio in the form of .
With further observation we realize that in order for the 2nd term to be , the first term has to be . So than in order to minimize the sum, we minimize have to .
The smallest possible value for such that it is an integer that's greater than is . So our first term is and our common ratio is . Thus the sum is or . Solution 2 by I_Dont_Do_Math (Solution edited by akaashp11)
See Also
2016 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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