2014 AMC 12A Problems/Problem 14

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 1

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)

Solution 2

Taking the definition of an arithmetic progression, there must be a common difference between the terms, giving us $(b-a) = (c-b)$ . From this, we can obtain the expression $a = 2b-c$ . Again, by taking the definition of a geometric progression, we can obtain the expression, $c=ar$ and $b=ar^2$ , where r serves as a value for the ratio between two terms in the progression. By substituting $b$ and $c$ in the arithmetic progression expression with the obtained values from the geometric progression, we obtain the equation, $a=2ar^2-ar$ which can be simplified to $(r-1)(2r+1)=0$ giving us $r=1$ of $r=-1/2$ . Thus, from the geometric progression, $a=a$ , $b=-1/2a$ and $c=1/4a$ . Looking at the initial conditions of $a<b<c$ we can see that $c=2$ or or $\boxed{\textbf{(C)}}$ is the lowest value provided that satisfied the above expressions.

2014 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2014 AMC 12A Problems/Problem 14

Contents

Problem

Solution 1

Solution 2

See Also