2004 AMC 12A Problems/Problem 14

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 1

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 $2(-14) + 29 = 1\Rightarrow\boxed{\mathrm{(A)}\ 1}$ .

Solution 2

Let $d$ be the common difference and $r$ be the common ratio. Then the arithmetic sequence is $9$ , $9+d$ , and $9+2d$ . The geometric sequence (when expressed in terms of $d$ ) has the terms $9$ , $11+d$ , and $29+2d$ . Thus, we get the following equations:

$9r=11+d\Rightarrow d=9r-11$

$9r^2=29+2d$

Plugging in the first equation into the second, our equation becomes $9r^2=29+18r-22\Longrightarrow9r^2-18r+7=0$ . By the quadratic formula, $r$ can either be $-\frac{1}{3}$ or $\frac{7}{3}$ . If $r$ is $-\frac{1}{3}$ , the third term (of the geometric sequence) would be $1$ , and if $r$ is $\frac{7}{3}$ , the third term would be $49$ . Clearly the minimum possible value for the third term of the geometric sequence is $\boxed{\mathrm{(A)}\ 1}$ .

2004 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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All AMC 12 Problems and Solutions

2004 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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All AMC 10 Problems and Solutions

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2004 AMC 12A Problems/Problem 14

Contents

Problem

Solution 1

Solution 2

See also