2004 AMC 10A Problems/Problem 18
Problem
A sequence of three real numbers forms an arithmetic progression with a first term of 9. If 2 is added to the secon term and 20 is added to the third term, the three resulting numbers form a geometric progression. What is the smallest possible value for the third term of the geometric progression?
Solution
Let d be the difference between terms in the arithmetic progression, such that first three terms are 9, 9+d, and 9+2d. The terms of the geometric progression will be 9, 11+d, and 29+2d. Because they are in a geometric progression, we can say
, note that the terms of a geometric progression would be , x, and , so .
Solve this equation.
Substituting into the geometric progression gives the terms 9, 21, and 49; substituting gives 9, -3, and 1. Therefore, the smallest possible 3rd term is 1 .
See also
2004 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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All AMC 10 Problems and Solutions |