2014 AMC 12A Problems/Problem 14
Contents
Problem
Let be three integers such that is an arithmetic progression and is a geometric progression. What is the smallest possible value of ?
Solution 1
We have , so . Since is geometric, . Since , we can't have and thus . Then our arithmetic progression is . Since , . The smallest possible value of is , or .
(Solution by AwesomeToad)
Solution 2
Taking the definition of an arithmetic progression, there must be a common difference between the terms, giving us . From this, we can obtain the expression . Again, by taking the definition of a geometric progression, we can obtain the expression, and , where r serves as a value for the ratio between two terms in the progression. By substituting and in the arithmetic progression expression with the obtained values from the geometric progression, we obtain the equation, which can be simplified to giving us of . Thus, from the geometric progression, , and . Looking at the initial conditions of we can see that the lowest integer value that would satisfy the above expressions is if , thus making or or
(Solution by thatuser)
See Also
2014 AMC 12A (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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