Difference between revisions of "2016 AMC 10B Problems/Problem 16"
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We know that the second term is the first term multiplied by the ratio. | We know that the second term is the first term multiplied by the ratio. | ||
In other words: | In other words: | ||
| − | <math>a_1 \cdot r=1</math> | + | <math>a_1 \cdot r= a_2</math>, |
| − | and | + | <math>a_2=1</math> (given), |
| − | <math>a_1=\frac{1}{r}</math> | + | <math>a_1 \cdot r=1</math>, and |
| + | <math>a_1=\frac{1}{r}</math>. | ||
Thus the sum is the following: | Thus the sum is the following: | ||
<math>S=\frac{\frac{1}{r}}{1-r}</math> | <math>S=\frac{\frac{1}{r}}{1-r}</math> | ||
Revision as of 13:10, 21 February 2016
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:
,
(given),
, and
.
Thus the sum is the following:
We can multiply to both sides of the numerator and denominator.
Since we want the minimum value of this expression, we want the maximum value for the denominator which is a quadratic of the form
The maximum value of a quadratic with negative 
is 
.
Plugging 1/2 in, we get:
, 
.
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. 
