Difference between revisions of "2016 AMC 12B Problems/Problem 14"
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==Solution 2== | ==Solution 2== | ||
+ | by Duseep Narala | ||
A simple approach is to initially recognize that <math>S_\infty = \frac{a}{1-r}</math> and <math>a=\frac{1}{r}</math>. We know that <math>|r| \leq 1</math>, since the series must converge. We can start by observing the greatest answer choice, 4. Therefore, <math>r\not\leq\frac{1}{3}</math>, because that would make <math>\frac{1}{r}\geq3</math>, which would make the series exceed 4. In order to minimize both the initial term and the rest of the series, we can recognize that <math>r=\frac{1}{2}</math> is the opitimal ratio, thus the answer is <math>\boxed{\textbf{(E)}\ 4}</math>. | A simple approach is to initially recognize that <math>S_\infty = \frac{a}{1-r}</math> and <math>a=\frac{1}{r}</math>. We know that <math>|r| \leq 1</math>, since the series must converge. We can start by observing the greatest answer choice, 4. Therefore, <math>r\not\leq\frac{1}{3}</math>, because that would make <math>\frac{1}{r}\geq3</math>, which would make the series exceed 4. In order to minimize both the initial term and the rest of the series, we can recognize that <math>r=\frac{1}{2}</math> is the opitimal ratio, thus the answer is <math>\boxed{\textbf{(E)}\ 4}</math>. |
Revision as of 15:55, 21 February 2016
Contents
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.
Calculus:
Let , then
. Since
and
are undefined
. This means that we only need to find where the derivative equals
, meaning
. So
, meaning that
AM-GM
For 2 positive real numbers and
,
. Let
and
. Then:
. This implies that
. or
. Rearranging :
. Thus, the smallest value is
.
Solution 2
by Duseep Narala
A simple approach is to initially recognize that and
. We know that
, since the series must converge. We can start by observing the greatest answer choice, 4. Therefore,
, because that would make
, which would make the series exceed 4. In order to minimize both the initial term and the rest of the series, we can recognize that
is the opitimal ratio, thus the answer is
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.