2016 AMC 10B Problems/Problem 16

Revision as of 14:08, 21 February 2016 by Shrummy (talk | contribs) (Solution)

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: S=\frac{a_1}{1-r}. 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: $a_1 \cdot r=1$ $a_1=\frac{1}{r}$ Thus the sum is the following: $S=\frac{\frac{1}{r}}{1-r}$ We can multiply $r$ to both sides of the numerator and denominator. $S=\frac{1}{r-r^2}$ Since we want the minimum value of this expression, we want the maximum value for the denominator. $max(-r^2+r)$ The maximum value of a quadratic with negative $a$ is $\frac{-b}{2a}$. $S=\frac{-(1)}{2(-1)}=\frac{1}{2}$ Plugging 1/2 in, we get: $S=\frac{1}{\frac{1}{2}}=2$, $B$.

See Also

2016 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AMC 10 Problems and Solutions

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