Difference between revisions of "2016 AMC 10B Problems/Problem 16"
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<math>S=\frac{a_1}{1-r}</math> | <math>S=\frac{a_1}{1-r}</math> | ||
where <math>a_1</math> is the first term and <math>r</math> is the ratio whose absolute value is less than 1. | where <math>a_1</math> is the first term and <math>r</math> is the ratio whose absolute value is less than 1. | ||
− | We know that the second term is the first term multiplied by the ratio. | + | We know that the second term(<math>1</math>) is the first term multiplied by the ratio. |
In other words: | In other words: | ||
<math>a_1 \cdot r= a_2</math>, | <math>a_1 \cdot r= a_2</math>, | ||
Line 21: | Line 21: | ||
<math>a_1=\frac{1}{r}</math>. | <math>a_1=\frac{1}{r}</math>. | ||
Thus the sum is the following: | Thus the sum is the following: | ||
− | <math>S=\frac{\frac{1}{r}}{1-r}</math> | + | <math>S=\frac{\frac{1}{r}}{1-r}</math>. |
We can multiply <math>r</math> to both sides of the numerator and denominator. | We can multiply <math>r</math> to both sides of the numerator and denominator. | ||
− | <math>S=\frac{1}{r-r^2}</math> | + | <math>S=\frac{1}{r-r^2}</math>. |
Since we want the minimum value of this expression, we want the maximum value for the denominator which is a quadratic of the form | Since we want the minimum value of this expression, we want the maximum value for the denominator which is a quadratic of the form | ||
− | <math>-r^2+r</math> | + | <math>-r^2+r</math>. |
The maximum value of a quadratic with negative <math>a</math> is <math>\frac{-b}{2a}</math>. | The maximum value of a quadratic with negative <math>a</math> is <math>\frac{-b}{2a}</math>. | ||
− | <math>S=\frac{-(1)}{2(-1)}=\frac{1}{2}</math> | + | <math>S=\frac{-(1)}{2(-1)}=\frac{1}{2}</math>. |
Plugging 1/2 in, we get: | Plugging 1/2 in, we get: | ||
<math>S=\frac{1}{\frac{1}{2}}=2</math>, <math>B</math>. | <math>S=\frac{1}{\frac{1}{2}}=2</math>, <math>B</math>. |
Revision as of 14:12, 21 February 2016
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
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:
,
(given),
, and
.
Thus the sum is the following:
.
We can multiply
to both sides of the numerator and denominator.
.
Since we want the minimum value of this expression, we want the maximum value for the denominator which is a quadratic of the form
.
The maximum value of a quadratic with negative
is
.
.
Plugging 1/2 in, we get:
,
.
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.