Difference between revisions of "2009 AMC 12A Problems/Problem 17"
(New page: == Problem == Let <math>a + ar_1 + ar_1^2 + ar_1^3 + \cdots</math> and <math>a + ar_2 + ar_2^2 + ar_2^3 + \cdots</math> be two different infinite geometric series of positive numbers with ...) |
The Anomaly (talk | contribs) (→Solution) |
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Using [[Vieta's formulas]] we get that the sum of these two roots is <math>\boxed{1}</math>. | Using [[Vieta's formulas]] we get that the sum of these two roots is <math>\boxed{1}</math>. | ||
+ | == Alternate Solution == | ||
+ | |||
+ | |||
+ | Using the formula for the sum of a geometric series we get that the sums of the given two sequences are <math>\frac a{1-r_1}</math> and <math>\frac a{1-r_2}</math>. | ||
+ | |||
+ | Hence we have <math>\frac a{1-r_1} = r_1</math> and <math>\frac a{1-r_2} = r_2</math>. | ||
+ | This can be rewritten as <math>r_1(1-r_1) = r_2(1-r_2) = a</math>. | ||
+ | |||
+ | Which can be further rewritten as <math>r_1-r_1^2 = r_2-r_2^2</math>. | ||
+ | Rearranging the equation we get <math>r_1-r_2 = r_1^2-r_2^2</math>. | ||
+ | Expressing this as a difference of squares we get <math> r_1-r_2 = (r_1-r_2)(r_1+r_2)</math>. | ||
+ | |||
+ | Dividing by like terms we finally get <math>r_1+r_2 = \boxed{1}</math> as desired. | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2009|ab=A|num-b=16|num-a=18}} | {{AMC12 box|year=2009|ab=A|num-b=16|num-a=18}} |
Revision as of 00:36, 19 February 2009
Contents
[hide]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 .
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.
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 |