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2007 AMC 12B Problems/Problem 15

Revision as of 14:19, 20 February 2008 by Chickendude (talk | contribs) (New page: ==Problem 15== The geometric series <math>a+ar+ar^2\cdots</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 <ma...)

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Problem 15

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 $a+r$ ?

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

Solution

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

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

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

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

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

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

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

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