Difference between revisions of "2009 AMC 12A Problems/Problem 17"
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Dividing by like terms we finally get <math>r_1+r_2 = \boxed{1}</math> as desired. | Dividing by like terms we finally get <math>r_1+r_2 = \boxed{1}</math> as desired. | ||
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+ | Note: It is necessary to check that <math>r_1-r_2\ne 0</math>, as you cannot divide by zero. As the problem states that the series are different, <math>r_1 \ne r_2</math>, and so there is no division by zero error. | ||
== 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 23:49, 2 July 2013
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.
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 |