Difference between revisions of "2014 AMC 12A Problems/Problem 14"

Revision as of 14:33, 8 February 2014

Problem

Let $a<b<c$ be three integers such that $a,b,c$ is an arithmetic progression and $a,c,b$ is a geometric progression. What is the smallest possible value of $c$ ?

$\textbf{(A) }-2\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }4\qquad \textbf{(E) }6\qquad$

Solution

We have $b-a=c-b$ , so $a=2b-c$ . Since $a,c,b$ is geometric, $c^2=ab=(2b-c)b \Rightarrow 2b^2-bc-c^2=(2b+c)(b-c)=0$ . Since $a<b<c$ , we can't have $b=c$ and thus $c=-2b$ . then our arithmetic progression is $4b,b,-2b$ . Since $4b < b < -2b$ , $b < 0$ . The smallest possible value of $c=-2b$ is $(-2)(-1)=2$ , or $\boxed{\textbf{(C)}}$ .

(Solution by AwesomeToad)

2014 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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All AMC 12 Problems and Solutions

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

@@ Line 14: / Line 14: @@
 (Solution by AwesomeToad)
+==See Also==
+{{AMC12 box|year=2014|ab=A|num-b=13|num-a=15}}
+{{MAA Notice}}

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Difference between revisions of "2014 AMC 12A Problems/Problem 14"

Revision as of 14:33, 8 February 2014

Problem

Solution

See Also