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Difference between revisions of "2016 AMC 10B Problems/Problem 16"

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==Solution==
 
==Solution==
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The sum of an infinite geometric series is a positive number <math>S</math>, and the second term in the series is <math>1</math>. What is the smallest possible value of <math>S?</math>
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The sum of an infinite geometric series is of the form:
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<math>(a1/(1-r)</math> where <math>a1</math> is the first term and <math>r</math> is the ratio whose absolute value is less than 1.
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We know that the second term is the first term multiplied by the ratio.
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In other words:
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<math>a1*r=1</math>
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<math>a1=1/r</math>
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Thus the sum is the following:
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<math>(1/r)/(1-r)</math>
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We can multiply <math>r</math> to both sides of the numerator and denominator.
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<math>1/(r-r^2)</math>
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Since we want the minimum value of this expression, we want the maximum value for the denominator.
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<math>max(-r^2+r)</math>
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The maximum value of this is <math>-b/2a</math>.
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<math>-(1)/2(-1)=1/2</math>
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Plugging 1/2 in, we get:
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<math>1/(1/2)=2</math>, <math>B</math>
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2016|ab=B|num-b=15|num-a=17}}
 
{{AMC10 box|year=2016|ab=B|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:34, 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 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: $1/(1/2)=2$, $B$

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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