Difference between revisions of "2014 AMC 12A Problems/Problem 14"
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\textbf{(E) }6\qquad</math> | \textbf{(E) }6\qquad</math> | ||
− | ==Solution== | + | ==Solution 1== |
We have <math>b-a=c-b</math>, so <math>a=2b-c</math>. Since <math>a,c,b</math> is geometric, <math>c^2=ab=(2b-c)b \Rightarrow 2b^2-bc-c^2=(2b+c)(b-c)=0</math>. Since <math>a<b<c</math>, we can't have <math>b=c</math> and thus <math>c=-2b</math>. Then our arithmetic progression is <math>4b,b,-2b</math>. Since <math>4b < b < -2b</math>, <math>b < 0</math>. The smallest possible value of <math>c=-2b</math> is <math>(-2)(-1)=2</math>, or <math>\boxed{\textbf{(C)}}</math>. | We have <math>b-a=c-b</math>, so <math>a=2b-c</math>. Since <math>a,c,b</math> is geometric, <math>c^2=ab=(2b-c)b \Rightarrow 2b^2-bc-c^2=(2b+c)(b-c)=0</math>. Since <math>a<b<c</math>, we can't have <math>b=c</math> and thus <math>c=-2b</math>. Then our arithmetic progression is <math>4b,b,-2b</math>. Since <math>4b < b < -2b</math>, <math>b < 0</math>. The smallest possible value of <math>c=-2b</math> is <math>(-2)(-1)=2</math>, or <math>\boxed{\textbf{(C)}}</math>. | ||
(Solution by AwesomeToad) | (Solution by AwesomeToad) | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Taking the definition of an arithmetic progression, there must be a common difference between the terms, giving us <math>(b-a) = (c-b)</math>. From this, we can obtain the expression <math>a = 2b-c</math>. Again, by taking the definition of a geometric progression, we can obtain the expression, <math>c=ar</math> and <math>b=ar^2</math>, where r serves as a value for the ratio between two terms in the progression. By substituting <math>b</math> and <math>c</math> in the arithmetic progression expression with the obtained values from the geometric progression, we obtain the equation, <math>a=2ar^2-ar</math> which can be simplified to <math>(r-1)(2r+1)=0</math> giving us <math>r=1</math> of <math>r=-1/2</math>. Thus, from the geometric progression, <math>a=a</math>, <math>b=-1/2a</math> and <math>c=1/4a</math>. Looking at the initial conditions of <math>a<b<c</math> we can see that <math>c=2</math> or or <math>\boxed{\textbf{(C)}}</math> is the lowest value provided that satisfied the above expressions. | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2014|ab=A|num-b=13|num-a=15}} | {{AMC12 box|year=2014|ab=A|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 11:10, 22 February 2015
Contents
[hide]Problem
Let be three integers such that
is an arithmetic progression and
is a geometric progression. What is the smallest possible value of
?
Solution 1
We have , so
. Since
is geometric,
. Since
, we can't have
and thus
. Then our arithmetic progression is
. Since
,
. The smallest possible value of
is
, or
.
(Solution by AwesomeToad)
Solution 2
Taking the definition of an arithmetic progression, there must be a common difference between the terms, giving us . From this, we can obtain the expression
. Again, by taking the definition of a geometric progression, we can obtain the expression,
and
, where r serves as a value for the ratio between two terms in the progression. By substituting
and
in the arithmetic progression expression with the obtained values from the geometric progression, we obtain the equation,
which can be simplified to
giving us
of
. Thus, from the geometric progression,
,
and
. Looking at the initial conditions of
we can see that
or or
is the lowest value provided that satisfied the above expressions.
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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