2016 AMC 12B Problem 14

Below is the professionally curated solution for Problem 14 of the 2016 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2016 AMC 12B solutions, or check the answer key.

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Concepts:geometric sequence quadratic optimization

Difficulty rating: 1730

14.

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 $S?$

$\dfrac{1+\sqrt5}{2}$

$2$

$\sqrt5$

$3$

$4$

Solution:

Let $r$ be the common ratio. Since the second term is $1,$ the first term is $\dfrac1r,$ so $S=\dfrac{1/r}{1-r}=\dfrac{1}{r-r^2}.$ Because $S\gt0,$ it is smallest when $r-r^2$ is largest. The parabola $r-r^2$ peaks at $r=\tfrac12,$ where it equals $\tfrac14,$ so the smallest value of $S$ is $\dfrac{1}{1/4}=4.$

Thus, the correct answer is E.

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