2016 AMC 10B Problems/Problem 16

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 a positive number $S$ , and the second term in the series is $1$ . What is the smallest possible value of $S?$ The sum of an infinite geometric series is of the form: $(a1/(1-r)$ where $a1$ 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: $a1*r=1$ $a1=1/r$ Thus the sum is the following: $(1/r)/(1-r)$ We can multiply $r$ to both sides of the numerator and denominator. $1/(r-r^2)$ 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 this is $-b/2a$ . $-(1)/2(-1)=1/2$ Plugging 1/2 in, we get: $1/(1/2)=2$ , $B$

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
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2016 AMC 10B Problems/Problem 16

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