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

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

Latest revision as of 20:42, 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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