2016 AMC 12B Problems/Problem 14
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: $S_\infinit = \frac{a}{1-r}$ (Error compiling LaTeX. Unknown error_msg) and substituting in 
, we get that $S_\infint = \frac{\frac{1}{r}}{1-r} = \frac{1}{1-r^2}$ (Error compiling LaTeX. Unknown error_msg).
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 | |
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