2004 AMC 10A Problems/Problem 18

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Problem

A sequence of three real numbers forms an arithmetic progression with a first term of 9. If 2 is added to the secon 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?

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

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

$(11+d)^2=9(29+2d)$, note that the terms of a geometric progression would be $\frac{x}r$, x, and $xr$, so $x^2=\frac{x}r\times r$.

Solve this equation.

$d^2+4d-140=0$

$(d-10)(d+14)=0$

Substituting $d=10$ into the geometric progression gives the terms 9, 21, and 49; substituting $d=14$ gives 9, -3, and 1. Therefore, the smallest possible 3rd term is 1 $\Rightarrow\mathrm{(A)}$.

See also

2004 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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All AMC 10 Problems and Solutions
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