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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 tr=1</math>
+
   <math>a_1 \cdot r=1</math>
 
   <math>a_1=\frac{1}{r}</math>
 
   <math>a_1=\frac{1}{r}</math>
 
Thus the sum is the following:
 
Thus the sum is the following:

Revision as of 13:00, 21 February 2016

Problem

The sum of an infinite geometric series is a positive number $S$, and the second term in the series is $1$. What is the smallest possible value of $S?$

$\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$


Solution

The sum of an infinite geometric series is of the form: 

  $S=\frac{a_1}{1-r}$

where $a_1$ is the first term and $r$ 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: 

  $a_1 \cdot r=1$
  $a_1=\frac{1}{r}$

Thus the sum is the following: 

  $S=\frac{\frac{1}{r}}{1-r}$

We can multiply $r$ to both sides of the numerator and denominator. 

  $S=\frac{1}{r-r^2}$

Since we want the minimum value of this expression, we want the maximum value for the denominator. 

  $max(-r^2+r)$

The maximum value of a quadratic with negative $a$ is $\frac{-b}{2a}$. 

 $S=\frac{-(1)}{2(-1)}=\frac{1}{2}$

Plugging 1/2 in, we get: 

 $S=\frac{1}{\frac{1}{2}}=2$, $B$

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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