Difference between revisions of "2014 AMC 12A Problems/Problem 14"
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By the definition of an arithmetic progression, we can label the terms <math>a</math>, <math>b</math>, and <math>c</math>, as <math>a</math>, <math>a+d</math>, and <math>a+2d</math>. Now, we have that <math>a</math>, <math>a+2d</math>, and <math>a+d</math> form a geometric progression. Since a geometric ratio has a common ratio between terms, we have <math>(a+2d)/a = (a+d)/a+2d</math>. Cross multiplying, we obtain the equation <math>(a+2d)^2=a(a+d)</math>. Multiplying it out and cancelling terms, we are left with the quadratic equation <math>4d^2+3ad=0</math>. Solving for <math>d</math> in terms of <math>a</math>, we get that <math>d=-3a/4</math> or <math>d=0</math>. Looking at the problem, we know that the <math>d</math> cannot be 0, therefore the arithmetic progression is <math>a, a/4, -a/2</math>, so we need to find the minimum value of <math>-a/2</math> while <math>-a/2>a</math>. Looking at our progression, we realize that a must be a multiple of 4 so that every term is an integer. Substituting <math>a=-4</math>, since that would yield the smallest value of <math>-a/2</math> which satisfies the conditions, we figure out that the answer is <math>\boxed{\textbf{(C)}}</math>. | By the definition of an arithmetic progression, we can label the terms <math>a</math>, <math>b</math>, and <math>c</math>, as <math>a</math>, <math>a+d</math>, and <math>a+2d</math>. Now, we have that <math>a</math>, <math>a+2d</math>, and <math>a+d</math> form a geometric progression. Since a geometric ratio has a common ratio between terms, we have <math>(a+2d)/a = (a+d)/a+2d</math>. Cross multiplying, we obtain the equation <math>(a+2d)^2=a(a+d)</math>. Multiplying it out and cancelling terms, we are left with the quadratic equation <math>4d^2+3ad=0</math>. Solving for <math>d</math> in terms of <math>a</math>, we get that <math>d=-3a/4</math> or <math>d=0</math>. Looking at the problem, we know that the <math>d</math> cannot be 0, therefore the arithmetic progression is <math>a, a/4, -a/2</math>, so we need to find the minimum value of <math>-a/2</math> while <math>-a/2>a</math>. Looking at our progression, we realize that a must be a multiple of 4 so that every term is an integer. Substituting <math>a=-4</math>, since that would yield the smallest value of <math>-a/2</math> which satisfies the conditions, we figure out that the answer is <math>\boxed{\textbf{(C)}}</math>. | ||
(Solution by i8Pie) | (Solution by i8Pie) | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/sxBuSZGfmHg | ||
+ | |||
+ | ~Lucas | ||
==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}} |
Latest revision as of 17:25, 19 September 2022
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 AT)
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
or
. Thus, from the geometric progression,
,
and
. Looking at the initial conditions of
we can see that the lowest integer value that would satisfy the above expressions is if
, thus making
or
(Solution by thatuser)
Solution 3
By the definition of an arithmetic progression, we can label the terms ,
, and
, as
,
, and
. Now, we have that
,
, and
form a geometric progression. Since a geometric ratio has a common ratio between terms, we have
. Cross multiplying, we obtain the equation
. Multiplying it out and cancelling terms, we are left with the quadratic equation
. Solving for
in terms of
, we get that
or
. Looking at the problem, we know that the
cannot be 0, therefore the arithmetic progression is
, so we need to find the minimum value of
while
. Looking at our progression, we realize that a must be a multiple of 4 so that every term is an integer. Substituting
, since that would yield the smallest value of
which satisfies the conditions, we figure out that the answer is
.
(Solution by i8Pie)
Video Solution
~Lucas
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.