Difference between revisions of "2016 AMC 10B Problems/Problem 16"
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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>. | 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>. | ||
− | + | Because the first term has to be <math>x</math>. So than in order to minimize the sum, we have minimize 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>. | 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>. |
Revision as of 16:32, 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 by akaashp11)
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 .
Because the first term has to be . So than in order to minimize the sum, we have minimize 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
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 |
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