2014 AMC 12A Problem 14

Attempt Problem 14 of the 2014 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2014 AMC 12A solutions, or check the answer key.

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14.

Let

a\lt b\lt 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 for

c?

-2

1

2

4

6

Let the common difference be

d

b=a+d

and

c=a+2d

Since

a,c,b

is geometric,

c^2=ab

; this simplifies to

3a+4d=0

Let

d=b-a\gt0,

b=a+d

and

c=a+2d.

Since

a,c,b

is geometric,

\begin{gathered} \dfrac{c}{a}=\dfrac{b}{c}\\ \implies(a+2d)^2=a(a+d), \end{gathered}

which simplifies to

3ad+4d^2=0,

3a+4d=0.

Then

a=-4k

and

d=3k

for a positive integer

k,

giving

c=a+2d=2k.

The smallest value is

c=2

(with

a=-4,

b=-1,

c=2

Thus, the correct answer is C.

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