2016 AMC 10B Problems/Problem 16
Contents
Problem
The sum of an infinite geometric series is a positive number , and the second term in the series is
. What is the smallest possible value of
Solution 1
The sum of an infinite geometric series is of the form:
where
is the first term and
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:
Thus, the sum is the following:
Since we want the minimum value of this expression, we want the maximum value for the denominator,
.
The maximum x-value of a quadratic with negative
is
.
Plugging
into the quadratic yields:
Therefore, the minimum sum of our infinite geometric sequence is .
(Solution by akaashp11)
As an extension to find the maximum value for the denominator we can find the derivative of
to get
. we know that this changes sign when r =
so plugging it in into the original equation we find the answer is
.
Solution 2
After observation we realize that in order to minimize our sum with
being the reciprocal of r. The common ratio
has to be in the form of
with
being an integer as anything more than
divided by
would give a larger sum than a ratio in the form of
.
The first term has to be , so then in order to minimize the sum, we have minimize
.
The smallest possible value for such that it is an integer that's greater than
is
. So our first term is
and our common ratio is
. Thus the sum is
or
.
Solution 2 by No_One
Solution 3
We can see that if is the first term, and
is the common ratio between each of the terms, then we can get
Also, we know that the second term can be expressed as
notice if we multiply
by
, we would get
This quadratic has real solutions if the discriminant is greater than or equal to zero, or
This yields that
or
.
However, since we know that
has to be positive, we can safely conclude that the minimum possible value of
is
.
Solution 4 (Quick Method)
Let the first term of the geometric series . Since it must be decreasing, we have
and the third term is
. Realize that by AM-GM inequality
with equality if
. However, we established that
so that means
. So the sum of the first three terms of the sequence
is greater than
, and the geometric series keeps continuing infinitely. This means the sum continues increasing. The only answer choice greater than
is
. ~skyscraper
Solution 5 (Clever Algebra)
Let the first term be The sum of the series is
Rewrite this as
By AM-GM we know that
so the minimum is
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See Also
2016 AMC 10B (Problems • Answer Key • Resources) | ||
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 | ||
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