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Difference between revisions of "2002 AMC 12B Problems/Problem 9"

(New page: The answer is 1/4. The arithmetic sequence doesn't require much thought as it is 1,2,3,4. The geometric sequence is 1,2,4.)
 
m (→‎Solution 1)
 
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The answer is 1/4.
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==Problem==
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If <math>a,b,c,d</math> are positive real numbers such that <math>a,b,c,d</math> form an increasing arithmetic sequence and <math>a,b,d</math> form a geometric sequence, then <math>\frac ad</math> is
  
−
The arithmetic sequence doesn't require much thought as it is 1,2,3,4.
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<math>\mathrm{(A)}\ \frac 1{12}
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\qquad\mathrm{(B)}\ \frac 16
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\qquad\mathrm{(C)}\ \frac 14
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\qquad\mathrm{(D)}\ \frac 13
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\qquad\mathrm{(E)}\ \frac 12</math>
  
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The geometric sequence is 1,2,4.
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==Solution==
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=== Solution 1 ===
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We can let <math>a=1</math>, <math>b=2</math>, <math>c=3</math>, and <math>d=4</math>. <math>\frac{a}{d}=\boxed{\frac{1}{4}}  \Longrightarrow \mathrm{(C)}</math>
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=== Solution 2 ===
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As <math>a, b, d</math> is a geometric sequence, let <math>b=ka</math> and <math>d=k^2a</math> for some <math>k>0</math>.
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Now, <math>a, b, c, d</math> is an arithmetic sequence. Its difference is <math>b-a=(k-1)a</math>. Thus <math>d=a + 3(k-1)a = (3k-2)a</math>.
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Comparing the two expressions for <math>d</math> we get <math>k^2=3k-2</math>. The positive solution is <math>k=2</math>, and <math>\frac{a}{d}=\frac{a}{k^2a}=\frac{1}{k^2}=\boxed{\frac{1}{4}} \Longrightarrow \mathrm{(C)}</math>.
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=== Solution 3 ===
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Letting <math>n</math> be the common difference of the arithmetic progression, we have
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<math>b = a + n</math>, <math>c = a + 2n</math>, <math>d = a + 3n</math>. We are given that <math>b / a</math> = <math>d / b</math>, or
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<cmath>\frac{a + n}{a} = \frac{a + 3n}{a + n}.</cmath>
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Cross-multiplying, we get
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<cmath>a^2 + 2an + n^2 = a^2 + 3an</cmath>
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<cmath>n^2 = an</cmath>
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<cmath>n = a</cmath>
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So <math>\frac{a}{d} = \frac{a}{a + 3n} = \frac{a}{4a} = \boxed{\frac{1}{4}}  \Longrightarrow \mathrm{(C)}</math>.
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==See also==
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{{AMC12 box|year=2002|ab=B|num-b=8|num-a=10}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 21:42, 11 December 2017

Problem

If $a,b,c,d$ are positive real numbers such that $a,b,c,d$ form an increasing arithmetic sequence and $a,b,d$ form a geometric sequence, then $\frac ad$ is

$\mathrm{(A)}\ \frac 1{12} \qquad\mathrm{(B)}\ \frac 16 \qquad\mathrm{(C)}\ \frac 14 \qquad\mathrm{(D)}\ \frac 13 \qquad\mathrm{(E)}\ \frac 12$

Solution

Solution 1

We can let $a=1$, $b=2$, $c=3$, and $d=4$. $\frac{a}{d}=\boxed{\frac{1}{4}} \Longrightarrow \mathrm{(C)}$

Solution 2

As $a, b, d$ is a geometric sequence, let $b=ka$ and $d=k^2a$ for some $k>0$.

Now, $a, b, c, d$ is an arithmetic sequence. Its difference is $b-a=(k-1)a$. Thus $d=a + 3(k-1)a = (3k-2)a$.

Comparing the two expressions for $d$ we get $k^2=3k-2$. The positive solution is $k=2$, and $\frac{a}{d}=\frac{a}{k^2a}=\frac{1}{k^2}=\boxed{\frac{1}{4}} \Longrightarrow \mathrm{(C)}$.

Solution 3

Letting $n$ be the common difference of the arithmetic progression, we have $b = a + n$, $c = a + 2n$, $d = a + 3n$. We are given that $b / a$ = $d / b$, or \[\frac{a + n}{a} = \frac{a + 3n}{a + n}.\] Cross-multiplying, we get \[a^2 + 2an + n^2 = a^2 + 3an\] \[n^2 = an\] \[n = a\] So $\frac{a}{d} = \frac{a}{a + 3n} = \frac{a}{4a} = \boxed{\frac{1}{4}} \Longrightarrow \mathrm{(C)}$.

See also

2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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

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