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

Revision as of 19:24, 11 December 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

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

@@ Line 52: / Line 52: @@
 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 solutions iff 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>.
 ==See Also==
 {{AMC10 box|year=2016|ab=B|num-b=15|num-a=17}}
 {{MAA Notice}}

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

Page

Toolbox

Search