Difference between revisions of "2016 AMC 10B Problems/Problem 16"
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Therefore, the minimum sum of our infinite geometric sequence is <math>\boxed{\textbf{(E)}\ 4}</math>. | Therefore, the minimum sum of our infinite geometric sequence is <math>\boxed{\textbf{(E)}\ 4}</math>. | ||
+ | ==Solution 2== | ||
+ | After observation we realize that in order to minimize our sum <math>\frac{a}{1-r}</math> with <math>a</math> being the reciprocal of r, the common ratio <math>r</math> has to be in the form of <math>1/x</math> with <math>x</math> being an integer as anything more than <math>1</math> divided by <math>x</math> would give a larger sum than a ratio in the form of <math>1/x</math>. | ||
+ | |||
+ | With further observation we realize that in order for the 2nd term to be <math>1</math>, the first term has to be <math>x</math>. So than in order to minimize the sum, we minimize have to <math>x</math>. | ||
+ | |||
+ | The smallest possible value for <math>x</math> such that it is an integer that's greater than <math>1</math> is <math>2</math>. So our first term is <math>2</math> and our common ratio is <math>1/2</math>. Thus the sum is <math>\frac{2}{1/2}</math> or <math>\boxed{\textbf{(E)}\ 4}</math>. | ||
+ | Solution 2 by I_Dont_Do_Math | ||
(Solution edited by akaashp11) | (Solution edited by akaashp11) | ||
Revision as of 14:56, 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 sum of an infinite geometric series is of the form:
where
is the first term and
is the ratio whose absolute value is less than 1.
We know that the second term is the first term multiplied by the ratio.
In other words:
Thus, the sum is the following:
Since we want the minimum value of this expression, we want the maximum value for the denominator,
.
The maximum x-value of a quadratic with negative
is
.
Plugging
into the quadratic yields:
Therefore, the minimum sum of our infinite geometric sequence is .
Solution 2
After observation we realize that in order to minimize our sum with
being the reciprocal of r, the common ratio
has to be in the form of
with
being an integer as anything more than
divided by
would give a larger sum than a ratio in the form of
.
With further observation we realize that in order for the 2nd term to be , the first term has to be
. So than in order to minimize the sum, we minimize have to
.
The smallest possible value for such that it is an integer that's greater than
is
. So our first term is
and our common ratio is
. Thus the sum is
or
.
Solution 2 by I_Dont_Do_Math
(Solution edited by akaashp11)
See Also
2016 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.