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> | ||
− | == Solution 1 == | + | == Solutions == |
+ | === Solution 1 === | ||
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>. | 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>. | ||
− | == Solution 2 == | + | === Solution 2 === |
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: | 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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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>. | 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>. | ||
− | ==Solution 3== | + | === Solution 3 === |
− | |||
Let the three numbers be, in increasing order, <math>z,y,9</math> | Let the three numbers be, in increasing order, <math>z,y,9</math> | ||
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Hence, we have that <math>9-y=y-z\implies 9+z=2y</math>. | Hence, we have that <math>9-y=y-z\implies 9+z=2y</math>. | ||
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Also, from the second part of information given, we get that | Also, from the second part of information given, we get that | ||
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<math>\frac{9}{y+2}=\frac{y+2}{z+20}\implies 9(z+20)=(y+2)^2\implies y=3(\sqrt{z+20})-2</math> | <math>\frac{9}{y+2}=\frac{y+2}{z+20}\implies 9(z+20)=(y+2)^2\implies y=3(\sqrt{z+20})-2</math> | ||
− | + | Plugging back in... | |
− | Plugging back in.. | ||
<math>9+z=6(\sqrt{z+20})-4\implies (9+z)^2=36(z+20)</math> | <math>9+z=6(\sqrt{z+20})-4\implies (9+z)^2=36(z+20)</math> | ||
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Simplifying, we get that <math>z^2-10z-551=0</math> | Simplifying, we get that <math>z^2-10z-551=0</math> | ||
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Hence, the answer is <math>-19+20=\boxed{1} \implies \boxed{A}</math> | Hence, the answer is <math>-19+20=\boxed{1} \implies \boxed{A}</math> | ||
+ | |||
+ | === Solution 4 === | ||
+ | Let the arithmetic sequence be <math>9,9+x,9+2x</math> and let the geometric sequence be <math>9,11+x,29+2x</math>. Now, we just try all the solutions. If the last term is <math>1</math>, then <math>x=-14</math>. This gives the geometric sequence <math>9,-3,1</math> which indeed works. The answer is <math>\boxed{1} \implies \boxed{A}</math> | ||
+ | |||
+ | Solution by franzliszt | ||
+ | |||
+ | === Solution 5 === | ||
+ | The terms of the arithmetic progression are 9, <math>9+d</math>, and <math>9+2d</math> for some real number <math>d</math>. The terms of the geometric progression are 9, <math>11+d</math>, and <math>29+2d</math>. Therefore<math> | ||
+ | (11+d)^{2} = 9(29+2d) \quad\text{so}\quad d^{2}+4d-140 = 0. | ||
+ | </math>Thus <math>d=10</math> or <math>d=-14</math>. The corresponding geometric progressions are <math>9, 21, 49</math> and <math>9, -3, 1,</math> so the smallest possible value for the third term of the geometric progression is <math>\boxed{1} \implies \boxed{A}</math>. | ||
+ | |||
+ | === Solution 6(Partially Brute Force) === | ||
+ | List out the first few terms arithmetic progressions and their corresponding geometric progressions. List both the positive and negative. Then you will find that the answer is <math>\boxed{A}</math>. | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/qy3ewAuOcSw | ||
+ | |||
+ | Education, the Study of Everything | ||
== See also == | == See also == |
Latest revision as of 11:08, 13 January 2021
- The following problem is from both the 2004 AMC 12A #14 and 2004 AMC 10A #18, so both problems redirect to this page.
Contents
Problem
A sequence of three real numbers forms an arithmetic progression with a first term of . If is added to the second term and 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?
Solutions
Solution 1
Let be the common difference. Then , , are the terms of the geometric progression. Since the middle term is the geometric mean of the other two terms, . The smallest possible value occurs when , and the third term is .
Solution 2
Let be the common difference and be the common ratio. Then the arithmetic sequence is , , and . The geometric sequence (when expressed in terms of ) has the terms , , and . Thus, we get the following equations:
Plugging in the first equation into the second, our equation becomes . By the quadratic formula, can either be or . If is , the third term (of the geometric sequence) would be , and if is , the third term would be . Clearly the minimum possible value for the third term of the geometric sequence is .
Solution 3
Let the three numbers be, in increasing order,
Hence, we have that .
Also, from the second part of information given, we get that
Plugging back in...
Simplifying, we get that
Applying the quadratic formula, we get that
Obviously, in order to minimize the value of , we have to subtract. Hence,
However, the problem asks for the minimum value of the third term in a geometric progression.
Hence, the answer is
Solution 4
Let the arithmetic sequence be and let the geometric sequence be . Now, we just try all the solutions. If the last term is , then . This gives the geometric sequence which indeed works. The answer is
Solution by franzliszt
Solution 5
The terms of the arithmetic progression are 9, , and for some real number . The terms of the geometric progression are 9, , and . ThereforeThus or . The corresponding geometric progressions are and so the smallest possible value for the third term of the geometric progression is .
Solution 6(Partially Brute Force)
List out the first few terms arithmetic progressions and their corresponding geometric progressions. List both the positive and negative. Then you will find that the answer is .
Video Solution
Education, the Study of Everything
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 |