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

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\textbf{(E)}\ 4</math>
 
\textbf{(E)}\ 4</math>
  
 +
==Solution 1==
 +
The sum of an infinite geometric series is of the form:
 +
<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.
  
−
==Solution==
 
−
The sum of an infinite geometric series is of the form:
 
−
<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.
 
 
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>a1*r=1</math>
+
<cmath>\begin{split}
−
<math>a1=1/r</math>
+
a_1 \cdot r & = 1 \\
−
Thus the sum is the following:
+
a_1 & = \frac{1}{r}
−
<math>(1/r)/(1-r)</math>
+
\end{split}</cmath>
−
We can multiply <math>r</math> to both sides of the numerator and denominator.
+
 
−
<math>1/(r-r^2)</math>
+
Thus, the sum is the following:
−
Since we want the minimum value of this expression, we want the maximum value for the denominator.
+
<cmath>\begin{split}
−
<math>max(-r^2+r)</math>
+
S & = \frac{\frac{1}{r}}{1-r} \\\\
−
The maximum value of this is <math>-b/2a</math>.
+
S & =\frac{1}{r-r^2}
−
<math>-(1)/2(-1)=1/2</math>
+
\end{split}</cmath>
−
Plugging 1/2 in, we get:
+
 
−
<math>1/(1/2)=2</math>, <math>B</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>.
 +
The maximum x-value of a quadratic with negative <math>a</math> is <math>\frac{-b}{2a}</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} -\left(\frac{1}{2}\right)^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>.
 +
(Solution by akaashp11)
 +
 
 +
As an extension to find the maximum value for the denominator we can find the derivative of <math>-r^2</math> <math>+</math> <math>r</math> to get <math>1</math> <math>-</math> <math>2r</math>. we know that this changes sign when r = <math>\frac{1}{2}</math> so plugging it in into the original equation we find the answer is <math>\boxed{\textbf{(E)}\ 4}</math>.
 +
 
 +
==Solution 2==
 +
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>.
 +
 
 +
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>.
 +
Solution 2 by No_One
 +
 
 +
==Solution 3==
 +
We can see that if <math>a</math> is the first term, and <math>r</math> is the common ratio between each of the terms, then we can get
 +
<cmath>S=\frac{a}{1-r} \implies S-Sr=a</cmath>
 +
Also, we know that the second term can be expressed as <math>a\cdot r</math>
 +
notice if we multiply <math>S-Sr=a</math> by <math>r</math>, we would get
 +
<cmath>r(S-Sr)=ar \implies Sr-Sr^2=1 \implies Sr^2-Sr+1=0</cmath>
 +
This quadratic has real solutions if the discriminant is greater than or equal to zero, or
 +
<cmath>S^2-4\cdot S \cdot 1 \ge 0</cmath>
 +
This yields that <math>S\le 0</math> or <math>S\ge 4</math>.
 +
However, since we know that <math>S</math> has to be positive, we can safely conclude that the minimum possible value of <math>S</math> is <math>\boxed{\textbf{(E)}\ 4}</math>.
 +
 
 +
==Solution 4 (Quick Method)==
 +
Let the first term of the geometric series <math>x</math>. Since it must be decreasing, we have <math>x>1</math> and the third term is <math>
 +
\frac{1}{x}</math>. Realize that by AM-GM inequality <math>x+\frac{1}{x} \ge 2</math> with equality iff <math>x = 1</math>. However, we established that <math>x>1</math> so that means <math>x+\frac{1}{x} > 2</math>. So the sum of the first three terms of the sequence <math>x + \frac{1}{x} + 1</math> is greater than <math>3</math>, and the geometric series keeps continuing  infinitely.  This means the sum continues increasing. The only answer choice greater than <math>3</math> is <math>\boxed{\textbf{(E)}\ 4}</math>. ~skyscraper
  
 
==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 19:27, 21 December 2019

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 1

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: \[\begin{split} S & = \frac{1}{\frac{1}{2} -\left(\frac{1}{2}\right)^2} \\\\ S & = \frac{1}{\frac{1}{4}} \end{split}\]

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

As an extension to find the maximum value for the denominator we can find the derivative of $-r^2$ $+$ $r$ to get $1$ $-$ $2r$. we know that this changes sign when r = $\frac{1}{2}$ so plugging it in into the original equation we find the answer is $\boxed{\textbf{(E)}\ 4}$.

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

Solution 3

We can see that if $a$ is the first term, and $r$ is the common ratio between each of the terms, then we can get \[S=\frac{a}{1-r} \implies S-Sr=a\] Also, we know that the second term can be expressed as $a\cdot r$ notice if we multiply $S-Sr=a$ by $r$, we would get \[r(S-Sr)=ar \implies Sr-Sr^2=1 \implies Sr^2-Sr+1=0\] This quadratic has real solutions if the discriminant is greater than or equal to zero, or \[S^2-4\cdot S \cdot 1 \ge 0\] This yields that $S\le 0$ or $S\ge 4$. However, since we know that $S$ has to be positive, we can safely conclude that the minimum possible value of $S$ is $\boxed{\textbf{(E)}\ 4}$.

Solution 4 (Quick Method)

Let the first term of the geometric series $x$. Since it must be decreasing, we have $x>1$ and the third term is $\frac{1}{x}$. Realize that by AM-GM inequality $x+\frac{1}{x} \ge 2$ with equality iff $x = 1$. However, we established that $x>1$ so that means $x+\frac{1}{x} > 2$. So the sum of the first three terms of the sequence $x + \frac{1}{x} + 1$ is greater than $3$, and the geometric series keeps continuing infinitely. This means the sum continues increasing. The only answer choice greater than $3$ is $\boxed{\textbf{(E)}\ 4}$. ~skyscraper

See Also

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