Difference between revisions of "2002 AMC 12B Problems/Problem 9"
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==Solution== | ==Solution== | ||
=== Solution 1 === | === 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 === | === Solution 2 === | ||
Line 25: | Line 25: | ||
<cmath>\frac{a + n}{a} = \frac{a + 3n}{a + n}.</cmath> | <cmath>\frac{a + n}{a} = \frac{a + 3n}{a + n}.</cmath> | ||
Cross-multiplying, we get | Cross-multiplying, we get | ||
− | <cmath>a^2 + 2an + n^2 | + | <cmath>a^2 + 2an + n^2 = a^2 + 3an</cmath> |
<cmath>n^2 = an</cmath> | <cmath>n^2 = an</cmath> | ||
<cmath>n = a</cmath> | <cmath>n = a</cmath> |
Latest revision as of 20:42, 11 December 2017
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 , , , and .
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 .
Solution 3
Letting be the common difference of the arithmetic progression, we have , , . We are given that = , or Cross-multiplying, we get So .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.