Difference between revisions of "2004 AMC 12A Problems/Problem 14"

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{{duplicate|[[2004 AMC 12A Problems|2004 AMC 12A #14]] and [[2004 AMC 10A Problems/Problem 18|2004 AMC 10A #18]]}}
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== Problem ==
 
== Problem ==
 
A [[sequence]] of three real numbers forms an [[arithmetic progression]] with a first term of <math>9</math>. If <math>2</math> is added to the second term and <math>20</math> 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?
 
A [[sequence]] of three real numbers forms an [[arithmetic progression]] with a first term of <math>9</math>. If <math>2</math> is added to the second term and <math>20</math> 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?
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== See also ==
 
== See also ==
 
{{AMC12 box|year=2004|ab=A|num-b=13|num-a=15}}
 
{{AMC12 box|year=2004|ab=A|num-b=13|num-a=15}}
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{{AMC10 box|year=2004|ab=A|num-b=17|num-a=19}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 15:16, 5 December 2007

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 $9$. If $2$ is added to the second 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 in the geometric progression?

$\text {(A)}\ 1 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 36 \qquad \text {(D)}\ 49 \qquad \text {(E)}\ 81$

Solution

Let $d$ be the common difference. Then $9, 9 + d + 2 = 11 + d, 9 + 2d + 20 = 29 + 2d$ are the terms of the geometric progression. Since the middle term is the geometric mean of the other two terms, $(11+d)^2 = 9(2d+29)$ $\Longrightarrow d^2 + 4d - 140$ $= (d+14)(d-10) = 0$. The smallest possible value occurs when $d = -14$, and the third term is $2(-14) + 29 = 1\ \mathrm{(A)}$.

See also

2004 AMC 12A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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