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

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

Revision as of 08:49, 5 February 2008

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

We can let a=1, b=2, c=3, and d=4. $\frac{a}{d}=\frac{1}{4}  \Rightarrow \boxed{\mathrm{(C)}}$

See also