Difference between revisions of "2004 AMC 12A Problems/Problem 14"
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<math>9r^2=29+2d</math> | <math>9r^2=29+2d</math> | ||
− | Plugging in the first equation into the second, our equation becomes <math>9r^2=29+18r-22\Longrightarrow9r^2-18r | + | Plugging in the first equation into the second, our equation becomes <math>9r^2=29+18r-22\Longrightarrow9r^2-18r+7=0</math>. By the quadratic formula, <math>r</math> can either be <math>-\frac{1}{3}</math> or <math>\frac{7}{3}</math>. If <math>r</math> is <math>-\frac{1}{3}</math>, the third term (of the geometric sequence) would be <math>1</math>, and if <math>r</math> is <math>\frac{7}{3}</math>, the third term would be <math>49</math>. Clearly the minimum possible value for the third term of the geometric sequence is <math>\boxed{\mathrm{(A)}\ 1}</math>. |
== See also == | == See also == |
Revision as of 11:04, 21 December 2016
- The following problem is from both the 2004 AMC 12A #14 and 2004 AMC 10A #18, so both problems redirect to this page.
Contents
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 1
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
.
Solution 2
Let be the common difference and
be the common ratio. Then the arithmetic sequence is
,
, and
. The geometric sequence (when expressed in terms of
) has the terms
,
, and
. Thus, we get the following equations:
Plugging in the first equation into the second, our equation becomes . By the quadratic formula,
can either be
or
. If
is
, the third term (of the geometric sequence) would be
, and if
is
, the third term would be
. Clearly the minimum possible value for the third term of the geometric sequence is
.
See also
2004 AMC 12A (Problems • Answer Key • Resources) | |
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 | |
All AMC 12 Problems and Solutions |
2004 AMC 10A (Problems • Answer Key • Resources) | ||
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 | ||
All AMC 10 Problems and Solutions |