Difference between revisions of "2016 AMC 12B Problems/Problem 14"

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AM-GM
 
AM-GM
For 2 positive real numbers <math>a</math> and <math>b</math>, <math>\frac{a+b}{2} \geq \sqrt{ab}</math>.
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For 2 positive real numbers <math>a</math> and <math>b</math>, <math>\frac{a+b}{2} \geq \sqrt{ab}</math>. Let <math>a = \frac{1}{r}</math> and <math>b = \frac{1}{1-r}</math>. Then: <math>\frac{\frac{1}{r}+\frac{1}{1-r}}{2} \geq \sqrt{\frac{1}{r}\cdot\frac{1}{1-r}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2016|ab=B|num-b=13|num-a=15}}
 
{{AMC12 box|year=2016|ab=B|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:26, 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 second term in a geometric series is $a_2 = a \cdot r$, where $r$ is the common ratio for the series and $a$ is the first term of the series. So we know that $a\cdot r = 1$ and we wish to find the minimum value of the infinite sum of the series. We know that: $S_\infty = \frac{a}{1-r}$ and substituting in $a=\frac{1}{r}$, we get that $S_\infty = \frac{\frac{1}{r}}{1-r} = \frac{1}{r(1-r)} = \frac{1}{r}+\frac{1}{1-r}$. From here, you can either use calculus or AM-GM.

Calculus: Let $f(x) = \frac{1}{x-x^2} = (x-x^2)^{-1}$, then $f'(x) = -(x-x^2)^{-2}\cdot (1-2x)$. Since $f(0)$ and $f(1)$ are undefined $x \neq 0,1$. This means that we only need to find where the derivative equals $0$, meaning $1-2x = 0 \Rightarrow x =\frac{1}{2}$. So $r = \frac{1}{2}$, meaning that $S_\infty = \frac{1}{\frac{1}{2} - (\frac{1}{2})^2} = \frac{1}{\frac{1}{2}-\frac{1}{4}} = \frac{1}{\frac{1}{4}} = 4$

AM-GM For 2 positive real numbers $a$ and $b$, $\frac{a+b}{2} \geq \sqrt{ab}$. Let $a = \frac{1}{r}$ and $b = \frac{1}{1-r}$. Then: $\frac{\frac{1}{r}+\frac{1}{1-r}}{2} \geq \sqrt{\frac{1}{r}\cdot\frac{1}{1-r}$ (Error compiling LaTeX. Unknown error_msg).

See Also

2016 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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 12 Problems and Solutions

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