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 AT)

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 $r=1$ or $r=-1/2$. Thus, from the geometric progression, $a=a$, $b=1/4a$ and $c=-1/2a$. Looking at the initial conditions of $a<b<c$ we can see that the lowest integer value that would satisfy the above expressions is if $a = -4$, thus making $c=2$ or $\boxed{\textbf{(C)}}$ (Solution by thatuser)

Solution 3

By the definition of an arithmetic progression, we can label the terms $a$, $b$, and $c$, as $a$, $a+d$, and $a+2d$. Now, we have that $a$, $a+2d$, and $a+d$ form a geometric progression. Since a geometric ratio has a common ratio between terms, we have $(a+2d)/a = (a+d)/a+2d$. Cross multiplying, we obtain the equation $(a+2d)^2=a(a+d)$. Multiplying it out and cancelling terms, we are left with the quadratic equation $4d^2+3ad=0$. Solving for $d$ in terms of $a$, we get that $d=-3a/4$ or $d=0$. Looking at the problem, we know that the $d$ cannot be 0, therefore the arithmetic progression is $a, a/4, -a/2$, so we need to find the minimum value of $-a/2$ while $-a/2>a$. Looking at our progression, we realize that a must be a multiple of 4 so that every term is an integer. Substituting $a=-4$, since that would yield the smallest value of $-a/2$ which satisfies the conditions, we figure out that the answer is $\boxed{\textbf{(C)}}$. (Solution by i8Pie)

See Also

2014 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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

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