2016 AMC 12B Problems/Problem 14

Revision as of 12:16, 21 February 2016 by Abvenkgoo (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 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 $S_\infty = \frac{\frac{1}{r}}{1-r} = \frac{1}{r(1-r)$ (Error compiling LaTeX. Unknown error_msg) = \frac{1}{r}+\frac{1}{1-r}$.

See Also

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

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