2004 AMC 12A Problems/Problem 14
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(Redirected from 2004 AMC 10A Problems/Problem 18)
- The following problem is from both the 2004 AMC 12A #14 and 2004 AMC 10A #18, so both problems redirect to this page.
Problem
A sequence of three real numbers forms an arithmetic progression with a first term of 
. If 
is added to the second term and 
is added to the third term, the three resulting numbers form a geometric progression. What is the smallest possible value for the third term in the geometric progression?
Solution
Let 
be the common difference. Then 
are the terms of the geometric progression. Since the middle term is the geometric mean of the other two terms, 
. The smallest possible value occurs when 
, and the third term is 
.
See also
| 2004 AMC 12A (Problems) | ||
| 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 | ||
| 2004 AMC 10A (Problems) | ||
| Preceded by Problem 17 | Followed by Problem 19 | |
| 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 | ||
