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2014 AMC 12A Problems/Problem 14

Revision as of 04:19, 8 September 2018 by Lolet (talk | contribs) (→‎Solution 1)

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) toijeaoijet

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, $a=a$, $b=1/4a$ 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)

See Also

2014 AMC 12A (Problems • Answer Key • Resources)
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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