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

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==Solution==
 
==Solution==
 
The sum of an infinite geometric series is of the form:
 
The sum of an infinite geometric series is of the form:
−
<math>S=\frac{a_1}{1-r}</math>
+
<cmath>\begin{split}
 +
S & = \frac{a_1}{1-r}
 +
\end{split}</cmath>
 
where <math>a_1</math> is the first term and <math>r</math> is the ratio whose absolute value is less than 1.
 
where <math>a_1</math> is the first term and <math>r</math> is the ratio whose absolute value is less than 1.
−
We know that the second term(<math>1</math>) 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 r= a_2</math>,
+
<cmath>\begin{split}
−
<math>a_2=1</math> (given),
+
a_1*r & = 1 \\
−
<math>a_1 \cdot r=1</math>, and
+
a_1 & = \frac{1}{r}
−
<math>a_1=\frac{1}{r}</math>.
+
\end{split}</cmath>
−
Thus the sum is the following:
+
 
−
<math>S=\frac{\frac{1}{r}}{1-r}</math>.
+
Thus, the sum is the following:
−
We can multiply <math>r</math> to both sides of the numerator and denominator.
+
<cmath>\begin{split}
−
<math>S=\frac{1}{r-r^2}</math>.
+
S & = \frac{\frac{1}{r}}{1-r} \\
−
Since we want the minimum value of this expression, we want the maximum value for the denominator which is a quadratic of the form
+
S & =\frac{1}{r-r^2}
−
<math>-r^2+r</math>.
+
\end{split}</cmath>
−
The maximum value of a quadratic with negative <math>a</math> is <math>\frac{-b}{2a}</math>.
+
 
−
<math>S=\frac{-(1)}{2(-1)}=\frac{1}{2}</math>.
+
Since we want the minimum value of this expression, we want the maximum value for the denominator, <math>-r^2</math> <math>+</math> <math>r</math>.
−
Plugging 1/2 in, we get:
+
The maximum x-value of a quadratic with negative <math>a</math> is <math>\frac{-b}{2a}</math>.
−
<math>S=\frac{1}{\frac{1}{2}}=2</math>, <math>B</math>.
+
<cmath>\begin{split}
 +
r & = \frac{-(1)}{2(-1)} \\
 +
r & = \frac{1}{2}
 +
\end{split}</cmath>
 +
 
 +
Plugging <math>r</math> <math>=</math> <math>\frac{1}{2}</math> into the quadratic yields:
 +
<cmath>\begin{split}
 +
S & = \frac{1}{\frac{1}{2} -(\frac{1}{2})^2} \\
 +
S & = \frac{1}{\frac{1}{4}}
 +
\end{split}</cmath>
 +
 
 +
Therefore, the minimum sum of our infinite geometric sequence is <math>\boxed{\textbf{(E)}\ 4}</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 13:15, 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: \[\begin{split} S & = \frac{a_1}{1-r} \end{split}\] 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: \[\begin{split} a_1*r & = 1 \\ a_1 & = \frac{1}{r} \end{split}\]

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 $\boxed{\textbf{(E)}\ 4}$.

See Also

2016 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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All AMC 10 Problems and Solutions

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