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

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 $\frac{a}{d} = \frac{a}{a + 3n} = \frac{a}{4a} = \boxed{\frac{1}{4}} \Longrightarrow \mathrm{(C)}$ .

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 10: / Line 10: @@
 ==Solution==
 === 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>
+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 ===
@@ Line 18: / Line 18: @@
 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>.
+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>.
+=== Solution 3 ===
+Letting <math>n</math> be the common difference of the arithmetic progression, we have
+<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
+<cmath>\frac{a + n}{a} = \frac{a + 3n}{a + n}.</cmath>
+Cross-multiplying, we get
+<cmath>a^2 + 2an + n^2 = a^2 + 3an</cmath>
+<cmath>n^2 = an</cmath>
+<cmath>n = a</cmath>
+So <math>\frac{a}{d} = \frac{a}{a + 3n} = \frac{a}{4a} = \boxed{\frac{1}{4}}  \Longrightarrow \mathrm{(C)}</math>.
 ==See also==
@@ Line 24: / Line 34: @@
 [[Category:Introductory Algebra Problems]]
+{{MAA Notice}}

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