Question

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 $S?$

$\ \dfrac{1+\sqrt{5}}{2}$

$\ 2$

$\ \sqrt{5}$

$\ 3$

$\ 4$

Answer: E

Concepts:geometric sequence completing the square optimization

Difficulty rating: 1540

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 .

2016 AMC 10B Problem 16

Answer: E

Difficulty rating: 1540

Solution:

Problem 16 in Other Years