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Difference between revisions of "2004 AMC 12A Problems/Problem 14"
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<math>\text {(A)}\ 1 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 36 \qquad \text {(D)}\ 49 \qquad \text {(E)}\ 81</math>
 
<math>\text {(A)}\ 1 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 36 \qquad \text {(D)}\ 49 \qquad \text {(E)}\ 81</math>
  
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== Solution ==
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== Solution 1 ==
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Let <math>d</math> be the common difference. Then <math>9, 9 + d + 2 = 11 + d, 9 + 2d + 20 = 29 + 2d</math> are the terms of the geometric progression. Since the middle term is the [[geometric mean]] of the other two terms, <math>(11+d)^2 = 9(2d+29)</math> <math>\Longrightarrow d^2 + 4d - 140</math> <math>= (d+14)(d-10) = 0</math>. The smallest possible value occurs when <math>d = -14</math>, and the third term is <math>2(-14) + 29 = 1\ \mathrm{(A)}</math>.
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Let <math>d</math> be the common difference. Then <math>9</math>, <math>9+d+2=11+d</math>, <math>9+2d+20=29+2d</math> are the terms of the geometric progression. Since the middle term is the [[geometric mean]] of the other two terms, <math>(11+d)^2 = 9(2d+29)</math> <math>\Longrightarrow d^2 + 4d - 140</math> <math>= (d+14)(d-10) = 0</math>. The smallest possible value occurs when <math>d = -14</math>, and the third term is <math>2(-14) + 29 = 1\Rightarrow\boxed{\mathrm{(A)}\ 1}</math>.
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== Solution 2 ==
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Let <math>d</math> be the common difference and <math>r</math> be the common ratio. Then the arithmetic sequence is <math>9</math>, <math>9+d</math>, and <math>9+2d</math>. The geometric sequence (when expressed in terms of <math>d</math>) has the terms <math>9</math>, <math>11+d</math>, and <math>29+2d</math>.  Thus, we get the following equations:
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<math>9r=11+d\Rightarrow d=9r-11</math>
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<math>9r^2=29+2d</math>
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Plugging in the first equation into the second, our equation becomes <math>9r^2=29+18r-22\Longrightarrow9r^2-18r-7=0</math>. By the quadratic formula, <math>r</math> can either be <math>-\frac{1}{3}</math> or <math>\frac{7}{3}</math>.  If <math>r</math> is <math>-\frac{1}{3}</math>, the third term (of the geometric sequence) would be <math>1</math>, and if <math>r</math> is <math>\frac{7}{3}</math>, the third term would be <math>49</math>.  Clearly the minimum possible value for the third term of the geometric sequence is <math>\boxed{\mathrm{(A)}\ 1}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 11:16, 21 July 2014

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 $\boxed{\mathrm{(A)}\ 1}$.

See also

2004 AMC 12A (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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
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