Difference between revisions of "2007 AMC 12B Problems/Problem 15"
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Clearly <math>(a + ar^2 + ar^4 + \ldots) = (ar + ar^3 + ar^5 + \ldots)/r = 3/r</math>. We obtain that <math>7 = 3/r + 3</math>, hence <math>r = \frac{3}{4}</math>. | Clearly <math>(a + ar^2 + ar^4 + \ldots) = (ar + ar^3 + ar^5 + \ldots)/r = 3/r</math>. We obtain that <math>7 = 3/r + 3</math>, hence <math>r = \frac{3}{4}</math>. | ||
− | Then from <math>7 = (a + ar + ar^2 + \ldots) = \frac{a}{1-r}</math> we get <math>a=\frac{7}{4}</math>, and thus <math>a + r = \frac{5}{2}</math>. | + | Then from <math>7 = (a + ar + ar^2 + \ldots) = \frac{a}{1-r}</math> we get <math>a=\frac{7}{4}</math>, and thus <math>a + r = \boxed{\frac{5}{2}}</math>. |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2007|ab=B|num-b=14|num-a=16}} | {{AMC12 box|year=2007|ab=B|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:07, 3 January 2020
Problem 15
The geometric series has a sum of , and the terms involving odd powers of have a sum of . What is ?
Solution
Solution 1
The sum of an infinite geometric series is given by where is the first term and is the common ratio.
In this series,
The series with odd powers of is given as
It's sum can be given by
Doing a little algebra
Solution 2
The given series can be decomposed as follows:
Clearly . We obtain that , hence .
Then from we get , and thus .
See Also
2007 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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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