2016 AMC 10B Problems/Problem 16

Revision as of 13:35, 21 February 2016 by Shrummy (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 sum of an infinite geometric series is of the form:

  $(a1/(1-r)$ where $a1$ is the first term and $r$ is the ratio whose absolute value is less than 1.

We know that the second term is the first term multiplied by the ratio. In other words:

  $a1*r=1$
  $a1=1/r$

Thus the sum is the following:

  $(1/r)/(1-r)$

We can multiply $r$ to both sides of the numerator and denominator.

  $1/(r-r^2)$

Since we want the minimum value of this expression, we want the maximum value for the denominator.

  $max(-r^2+r)$

The maximum value of a quadratic with negative $a$ is $-b/2a$.

 $-(1)/2(-1)=1/2$

Plugging 1/2 in, we get:

 $1/(1/2)=2$, $B$

See Also

2016 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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