2016 AMC 10B Problem 16

Below is the professionally curated solution for Problem 16 of the 2016 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2016 AMC 10B solutions, or check the answer key.

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Concepts:geometric sequencecompleting the squareoptimization

Difficulty rating: 1540

16.

The sum of an infinite geometric series is a positive number S,S, and the second term in the series is 1.1. What is the smallest possible value of S?S?

 1+52 \ \dfrac{1+\sqrt{5}}{2}

 2 \ 2

 5 \ \sqrt{5}

 3 \ 3

 4 \ 4

Solution:

Let the first value of the series be a,a, and let the ratio be r.r. Thus, S=a1r=arr(1r)=1r(1r).\begin{align*}S&=\dfrac{a}{1-r}\\ &= \dfrac{ar}{r(1-r)} \\&= \dfrac{1}{r(1-r)}.\end{align*} This means we have to find rr that maximizes r(1r)=0.25(r0.5)2.r(1-r)= 0.25-(r-0.5)^2. This maximization will happen with r=0.5.r=0.5.

Therefore, S=10.5(0.5)=4.S = \dfrac{1}{0.5(0.5)}= 4.

Thus, the correct answer is E .

Problem 16 in Other Years