Difference between revisions of "2002 AMC 12B Problems/Problem 9"
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==Solution== | ==Solution== | ||
− | We can let a=1, b=2, c=3, and d=4. <math>\frac{a}{d}=\frac{1}{4} \ | + | === 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>. | ||
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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}}</math>. | ||
==See also== | ==See also== |
Revision as of 15:20, 6 January 2009
Contents
[hide]Problem
If are positive real numbers such that form an increasing arithmetic sequence and form a geometric sequence, then is
Solution
Solution 1
We can let a=1, b=2, c=3, and d=4.
Solution 2
As is a geometric sequence, let and for some .
Now, is an arithmetic sequence. Its difference is . Thus .
Comparing the two expressions for we get . The positive solution is , and .
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 |