Difference between revisions of "2007 AMC 12B Problems/Problem 15"

Revision as of 00:19, 22 February 2008

The geometric series $a+ar+ar^2\cdots$ has a sum of $7$ , and the terms involving odd powers of $r$ have a sum of $3$ . What is $a+r$ ?

$\mathrm {(A)} \frac{4}{3}$ $\mathrm {(B)} \frac{12}{7}$ $\mathrm {(C)} \frac{3}{2}$ $\mathrm {(D)} \frac{7}{3}$ $\mathrm {(E)} \frac{5}{2}$

The sum of an infinite geometric series is given by $\frac{a}{1-r}$ where $a$ is the first term and $r$ is the common ratio.

In this series, $\frac{a}{1-r} = 7$

The series with odd powers of $r$ is given as $ar + ar^3 + ar^5 ...$

It's sum can be given by $\frac{ar}{1-r^2} = 3$

Doing a little algebra

$ar = 3(1-r)(1+r)$

$ar = 3\left(\frac{a}{7}\right)(1+r)$

$\frac{7}{3}r = 1 + r$

$r = \frac{3}{4}$

$a = 7(1-r) = \frac{7}{4}$

$a + r = \frac{5}{2} \Rightarrow \mathrm{(E)}$

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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 <math>a + r = \frac{5}{2} \Rightarrow \mathrm{(E)}</math>
+==See Also==
+{{AMC12 box|year=2007|ab=B|num-b=14|num-a=16}}