Difference between revisions of "2016 AMC 10B Problems/Problem 16"
(→Solution 1) |
(→Solution 1) |
||
Line 38: | Line 38: | ||
Plugging <math>r</math> <math>=</math> <math>\frac{1}{2}</math> into the quadratic yields: | Plugging <math>r</math> <math>=</math> <math>\frac{1}{2}</math> into the quadratic yields: | ||
<cmath>\begin{split} | <cmath>\begin{split} | ||
− | S & = \frac{1}{\frac{1}{2} -(\frac{1}{2})^2} \\ | + | S & = \frac{1}{\frac{1}{2} -\left(\frac{1}{2}\right)^2} \\ |
S & = \frac{1}{\frac{1}{4}} | S & = \frac{1}{\frac{1}{4}} | ||
\end{split}</cmath> | \end{split}</cmath> |
Revision as of 14:33, 29 January 2019
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 1
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 by akaashp11)
As an extension to find the maximum value for the denominator we can find the derivative of to get . we know that this changes sign when r = so plugging it in into the original equation we find the answer 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 .
The first term has to be , so then in order to minimize the sum, we have minimize .
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 No_One
Solution 3
We can see that if is the first term, and is the common ratio between each of the terms, then we can get Also, we know that the second term can be expressed as notice if we multiply by , we would get This quadratic has real solutions if the discriminant is greater than or equal to zero, or This yields that or . However, since we know that has to be positive, we can safely conclude that the minimum possible value of is .
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.