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−  ==Problem==
 +  #REDIRECT [[2004 AMC 12A Problems/Problem 14]] 
−  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 of the geometric progression?
 
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−  <math> \mathrm{(A) \ } 1 \qquad \mathrm{(B) \ } 4 \qquad \mathrm{(C) \ } 36 \qquad \mathrm{(D) \ } 49 \qquad \mathrm{(E) \ } 81 </math>
 
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−  ==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
 
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−  <math>(11+d)^2=9(29+2d)</math>, note that the terms of a geometric progression would be <math>\frac{x}r</math>, x, and <math>xr</math>, so <math>x^2=\frac{x}r\times r</math>.
 
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−  Solve this equation.
 
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−  <math>d^2+4d140=0</math>
 
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−  <math>(d10)(d+14)=0</math>
 
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−  Substituting <math>d=10</math> into the geometric progression gives the terms 9, 21, and 49; substituting <math>d=14</math> gives 9, 3, and 1. Therefore, the smallest possible 3rd term is 1 <math>\Rightarrow\mathrm{(A)}</math>.
 
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−  == See also ==
 
−  {{AMC10 boxyear=2004ab=Anumb=17numa=19}}
 
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−  [[Category:Introductory Algebra Problems]]  