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

Revision as of 15:20, 6 January 2009

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 $\frac{a}{d}=\frac{a}{k^2a}=\frac{1}{k^2}=\boxed{\frac{1}{4}}$ .

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

@@ Line 9: / Line 9: @@
 ==Solution==
-We can let a=1, b=2, c=3, and d=4. <math>\frac{a}{d}=\frac{1}{4}  \Rightarrow \boxed{\mathrm{(C)}}</math>
+=== Solution 1 ===
+We can let a=1, b=2, c=3, and d=4. <math>\frac{a}{d}=\boxed{\frac{1}{4}}  \Longrightarrow \mathrm{(C)}</math>
+=== Solution 2 ===
+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>.
+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>.
+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}}</math>.
 ==See also==

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