Difference between revisions of "2009 AMC 12A Problems/Problem 17"
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We basically have two infinite geometric series whose sum is equivalent to the common ratio. Let us have a geometric series: <math>b, br, br^2.....</math>. | We basically have two infinite geometric series whose sum is equivalent to the common ratio. Let us have a geometric series: <math>b, br, br^2.....</math>. | ||
− | The sum is: <math>\frac{b}{1-r} = r.</math> Thus, <math>b = r-r^2</math> and by Vieta's, the sum of the two possible values of <math>r (r_1 | + | The sum is: <math>\frac{b}{1-r} = r.</math> Thus, <math>b = r-r^2</math> and by Vieta's, the sum of the two possible values of <math>r</math> (<math>r_1</math> and <math>r_2</math>) is <math>1</math>. |
Revision as of 15:27, 31 July 2019
Problem
Let and
be two different infinite geometric series of positive numbers with the same first term. The sum of the first series is
, and the sum of the second series is
. What is
?
Solution
Using the formula for the sum of a geometric series we get that the sums of the given two sequences are and
.
Hence we have and
.
This can be rewritten as
.
As we are given that and
are distinct, these must be precisely the two roots of the equation
.
Using Vieta's formulas we get that the sum of these two roots is .
Solution 2
We basically have two infinite geometric series whose sum is equivalent to the common ratio. Let us have a geometric series: .
The sum is: Thus,
and by Vieta's, the sum of the two possible values of
(
and
) is
.
~conantwiz2023
Alternate Solution
Using the formula for the sum of a geometric series we get that the sums of the given two sequences are and
.
Hence we have and
.
This can be rewritten as
.
Which can be further rewritten as .
Rearranging the equation we get
.
Expressing this as a difference of squares we get
.
Dividing by like terms we finally get as desired.
Note: It is necessary to check that , as you cannot divide by zero. As the problem states that the series are different,
, and so there is no division by zero error.
See Also
2009 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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 12 Problems and Solutions |
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