2016 AMC 10B Problem 16

Problem 16 of 25IntermediateAlgebra

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?

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Solution

Let the first value of the series be

a,

and let the ratio be

r.

Thus,

\begin{aligned}S&=\dfrac{a}{1-r}\\ &= \dfrac{ar}{r(1-r)} \\&= \dfrac{1}{r(1-r)}.\end{aligned}

This means we have to find

r

that maximizes

r(1-r)= 0.25-(r-0.5)^2.

This maximization will happen with

r=0.5.

Therefore,

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

Thus, the correct answer is E .