Difference between revisions of "2002 AMC 12B Problems/Problem 9"
(New page: The answer is 1/4. The arithmetic sequence doesn't require much thought as it is 1,2,3,4. The geometric sequence is 1,2,4.) |
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| − | + | ==Problem== | |
| + | If <math>a,b,c,d</math> are positive real numbers such that <math>a,b,c,d</math> form an increasing arithmetic sequence and <math>a,b,d</math> form a geometric sequence, then <math>\frac ad</math> is | ||
| − | + | <math>\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</math> | ||
| − | + | ==Solution== | |
| + | We can let a=1, b=2, c=3, and d=4. <math>\frac{a}{d}=\frac{1}{4} -> \boxed{\mathrm{(C)}}</math> | ||
| + | |||
| + | ==See also== | ||
Revision as of 08:48, 5 February 2008
Problem
If 
are positive real numbers such that form an increasing arithmetic sequence and 
form a geometric sequence, then 
is
Solution
We can let a=1, b=2, c=3, and d=4. 
