Difference between revisions of "2007 AMC 12B Problems/Problem 15"
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− | ==Problem | + | ==Problem== |
The geometric series <math>a+ar+ar^2\ldots</math> has a sum of <math>7</math>, and the terms involving odd powers of <math>r</math> have a sum of <math>3</math>. What is <math>a+r</math>? | The geometric series <math>a+ar+ar^2\ldots</math> has a sum of <math>7</math>, and the terms involving odd powers of <math>r</math> have a sum of <math>3</math>. What is <math>a+r</math>? | ||
<math> \textbf{(A)}\ \frac{4}{3}\qquad\textbf{(B)}\ \frac{12}{7}\qquad\textbf{(C)}\ \frac{3}{2}\qquad\textbf{(D)}\ \frac{7}{3}\qquad\textbf{(E)}\ \frac{5}{2} </math> | <math> \textbf{(A)}\ \frac{4}{3}\qquad\textbf{(B)}\ \frac{12}{7}\qquad\textbf{(C)}\ \frac{3}{2}\qquad\textbf{(D)}\ \frac{7}{3}\qquad\textbf{(E)}\ \frac{5}{2} </math> | ||
− | + | ==Solution 1== | |
− | |||
The sum of an infinite geometric series is given by <math>\frac{a}{1-r}</math> where <math>a</math> is the first term and <math>r</math> is the common ratio. | The sum of an infinite geometric series is given by <math>\frac{a}{1-r}</math> where <math>a</math> is the first term and <math>r</math> is the common ratio. | ||
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The series with odd powers of <math>r</math> is given as <cmath>ar + ar^3 + ar^5 ...</cmath> | The series with odd powers of <math>r</math> is given as <cmath>ar + ar^3 + ar^5 ...</cmath> | ||
− | + | Its sum can be given by <math>\frac{ar}{1-r^2} = 3</math> | |
Doing a little algebra | Doing a little algebra | ||
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<math>a = 7(1-r) = \frac{7}{4}</math> | <math>a = 7(1-r) = \frac{7}{4}</math> | ||
− | <math>a + r = \frac{5}{2} \Rightarrow \mathrm{(E)}</math> | + | <math>a + r =\boxed{ \frac{5}{2}} \Rightarrow \mathrm{(E)}</math> |
− | + | ==Solution 2== | |
The given series can be decomposed as follows: <math>(a + ar + ar^2 + \ldots) = (a + ar^2 + ar^4 + \ldots) + (ar + ar^3 + ar^5 + \ldots)</math> | The given series can be decomposed as follows: <math>(a + ar + ar^2 + \ldots) = (a + ar^2 + ar^4 + \ldots) + (ar + ar^3 + ar^5 + \ldots)</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>. | 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}} |
Latest revision as of 15:41, 27 July 2021
Contents
Problem
The geometric series has a sum of
, and the terms involving odd powers of
have a sum of
. What is
?
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
Its 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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