2016 AMC 10B Problems/Problem 16
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:
![\[\begin{split} S & = \frac{a_1}{1-r} \end{split}\]](http://latex.artofproblemsolving.com/a/6/9/a697cd5bfeed3bf0b41ac850488992117247afd4.png)
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:
![\[\begin{split} a_1*r & = 1 \\ a_1 & = \frac{1}{r} \end{split}\]](http://latex.artofproblemsolving.com/6/5/b/65b2bafe134b0846f1790af04eb54245a5c59bb9.png)
Thus, the sum is the following:
![\[\begin{split} S & = \frac{\frac{1}{r}}{1-r} \\ S & =\frac{1}{r-r^2} \end{split}\]](http://latex.artofproblemsolving.com/7/7/1/7719ebd4ece0417d3141331146c9c7b0720a4b42.png)
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 
.
![\[\begin{split} r & = \frac{-(1)}{2(-1)} \\ r & = \frac{1}{2} \end{split}\]](http://latex.artofproblemsolving.com/1/e/1/1e1f45d005c38a4c2531ff4abb937e19ea3709a6.png)
Plugging 
into the quadratic yields:
![\[\begin{split} S & = \frac{1}{\frac{1}{2} -(\frac{1}{2})^2} \\ S & = \frac{1}{\frac{1}{4}} \end{split}\]](http://latex.artofproblemsolving.com/7/5/c/75cb96ac810cc58aa9547367e441e5d0e2faac23.png)
Therefore, the minimum sum of our infinite geometric sequence is 
.
(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 | ||
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