Difference between revisions of "2016 AMC 12B Problems/Problem 14"
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<cmath>S=\frac{a^2}{a-1}</cmath> | <cmath>S=\frac{a^2}{a-1}</cmath> | ||
− | We seek the smallest positive value of <math>S</math>. <math> | + | We seek the smallest positive value of <math>S</math>. <math>\frac{(a-2)a}{(a-1)^2}=S'</math> and <math>\frac{(a-2)a}{(a-1)^2}=0</math> at <math>a=0</math> and <math>a=2</math>. <math>\frac{2}{(a-1)^3}=S''</math> and <math>\frac{2}{(0-1)^3}</math> is negative (implying a maximum occurs at <math>a=0</math>) and <math>\frac{2}{(2-1)^3}</math> is positive (implying a minimum occurs at <math>a=2</math>). At <math>a=2</math>, <math>S=\boxed{\textbf{(E)}\ 4}.</math> This is the only positive minimum <math>S</math> has, so it must be the answer. |
<math>\textbf{Solving in terms of \textit{r} then being a clever cow}</math> | <math>\textbf{Solving in terms of \textit{r} then being a clever cow}</math> |
Revision as of 21:50, 5 August 2017
Contents
[hide]Problem
The sum of an infinite geometric series is a positive number , and the second term in the series is
. What is the smallest possible value of
Solution
The second term in a geometric series is , where
is the common ratio for the series and
is the first term of the series. So we know that
and we wish to find the minimum value of the infinite sum of the series. We know that:
and substituting in
, we get that
. From here, you can either use calculus or AM-GM.
Let , then
. Since
and
are undefined
. This means that we only need to find where the derivative equals
, meaning
. So
, meaning that
For 2 positive real numbers and
,
. Let
and
. Then:
. This implies that
. or
. Rearranging :
. Thus, the smallest value is
.
Solution 2
A geometric sequence always looks like
and they say that the second term . You should know that the sum of an infinite geometric series (denoted by
here) is
. We now have a system of equations which allows us to find
in one variable.
We seek the smallest positive value of . We proceed by graphing in the
plane and find the answer is
We seek the smallest positive value of . We proceed by graphing in the
plane and find the answer is
We seek the smallest positive value of .
and
at
and
.
and
is negative (implying a maximum occurs at
) and
is positive (implying a minimum occurs at
). At
,
This is the only positive minimum
has, so it must be the answer.
We seek the smallest positive value of . We could use calculus like we did in the solution immediately above this one, but that's a lot of work and we don't have a ton of time. To minimize the positive value of this fraction, we must maximize the denominator. The denominator is a quadratic that opens down, so its maximum occurs at its vertex. The vertex of this quadratic occurs at
and
.
See Also
2016 AMC 12B (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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