2014 AMC 12A Problems/Problem 14
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 AT)
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 or . 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 (Solution by thatuser)
Solution 3
By the definition of an arithmetic progression, we can label the terms , , and , as , , and . Now, we have that , , and form a geometric progression. Since a geometric ratio has a common ratio between terms, we have . Cross multiplying, we obtain the equation . Multiplying it out and cancelling terms, we are left with the quadratic equation . Solving for in terms of , we get that or . Looking at the problem, we know that the cannot be 0, therefore the arithmetic progression is , so we need to find the minimum value of while . Looking at our progression, we realize that a must be a multiple of 4 so that every term is an integer. Substituting , since that would yield the smallest value of which satisfies the conditions, we figure out that the answer is . (Solution by i8Pie)
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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