Difference between revisions of "2004 AMC 10A Problems/Problem 18"

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#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?
<math> \mathrm{(A) \ } 1 \qquad \mathrm{(B) \ } 4 \qquad \mathrm{(C) \ } 36 \qquad \mathrm{(D) \ } 49 \qquad \mathrm{(E) \ } 81  </math>
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
<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>.
Solve this equation.
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>.
== See also ==
{{AMC10 box|year=2004|ab=A|num-b=17|num-a=19}}
[[Category:Introductory Algebra Problems]]

Latest revision as of 16:15, 5 December 2007

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