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

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After observation we realize that in order to minimize our sum <math>\frac{a}{1-r}</math> with <math>a</math> being the reciprocal of r. The common ratio <math>r</math> has to be in the form of <math>1/x</math> with <math>x</math> being an integer as anything more than <math>1</math> divided by <math>x</math> would give a larger sum than a ratio in the form of <math>1/x</math>.  
 
After observation we realize that in order to minimize our sum <math>\frac{a}{1-r}</math> with <math>a</math> being the reciprocal of r. The common ratio <math>r</math> has to be in the form of <math>1/x</math> with <math>x</math> being an integer as anything more than <math>1</math> divided by <math>x</math> would give a larger sum than a ratio in the form of <math>1/x</math>.  
  
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The first term has to be <math>x</math>, so then in order to minim6tt5r5y,7uerthrtdhdthize the sum, we have minimize <math>x</math>.  
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The first term has to be <math>x</math>, so then in order to minimize the sum, we have minimize <math>x</math>.  
  
 
The smallest possible value for <math>x</math> such that it is an integer that's greater than <math>1</math> is <math>2</math>. So our first term is <math>2</math> and our common ratio is <math>1/2</math>. Thus the sum is <math>\frac{2}{1/2}</math> or <math>\boxed{\textbf{(E)}\ 4}</math>.
 
The smallest possible value for <math>x</math> such that it is an integer that's greater than <math>1</math> is <math>2</math>. So our first term is <math>2</math> and our common ratio is <math>1/2</math>. Thus the sum is <math>\frac{2}{1/2}</math> or <math>\boxed{\textbf{(E)}\ 4}</math>.

Revision as of 08:27, 5 October 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: \[\begin{split} S & = \frac{a_1}{1-r} \end{split}\] 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: \[\begin{split} a_1 \cdot r & = 1 \\ a_1 & = \frac{1}{r} \end{split}\]

Thus, the sum is the following: \[\begin{split} S & = \frac{\frac{1}{r}}{1-r} \\ S & =\frac{1}{r-r^2} \end{split}\]

Since we want the minimum value of this expression, we want the maximum value for the denominator, $-r^2$ $+$ $r$. The maximum x-value of a quadratic with negative $a$ is $\frac{-b}{2a}$. \[\begin{split} r & = \frac{-(1)}{2(-1)} \\ r & = \frac{1}{2} \end{split}\]

Plugging $r$ $=$ $\frac{1}{2}$ into the quadratic yields: \[\begin{split} S & = \frac{1}{\frac{1}{2} -(\frac{1}{2})^2} \\ S & = \frac{1}{\frac{1}{4}} \end{split}\]

Therefore, the minimum sum of our infinite geometric sequence is $\boxed{\textbf{(E)}\ 4}$. (Solution by akaashp11)

Solution 2

After observation we realize that in order to minimize our sum $\frac{a}{1-r}$ with $a$ being the reciprocal of r. The common ratio $r$ has to be in the form of $1/x$ with $x$ being an integer as anything more than $1$ divided by $x$ would give a larger sum than a ratio in the form of $1/x$.

The first term has to be $x$, so then in order to minimize the sum, we have minimize $x$.

The smallest possible value for $x$ such that it is an integer that's greater than $1$ is $2$. So our first term is $2$ and our common ratio is $1/2$. Thus the sum is $\frac{2}{1/2}$ or $\boxed{\textbf{(E)}\ 4}$. Solution 2 by No_One

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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