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Difference between revisions of "2007 AMC 12B Problems/Problem 15"
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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>
  
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It's sum can be given by <math>\frac{ar}{1-r^2} = 3</math>
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Its sum can be given by <math>\frac{ar}{1-r^2} = 3</math>
  
 
Doing a little algebra
 
Doing a little algebra

Latest revision as of 14:41, 27 July 2021

Problem

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

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

Solution 1

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 \[ar + ar^3 + ar^5 ...\]

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

Doing a little algebra

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

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

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

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

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

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

Solution 2

The given series can be decomposed as follows: $(a + ar + ar^2 + \ldots) = (a + ar^2 + ar^4 + \ldots) + (ar + ar^3 + ar^5 + \ldots)$

Clearly $(a + ar^2 + ar^4 + \ldots) = (ar + ar^3 + ar^5 + \ldots)/r = 3/r$. We obtain that $7 = 3/r + 3$, hence $r = \frac{3}{4}$.

Then from $7 = (a + ar + ar^2 + \ldots) = \frac{a}{1-r}$ we get $a=\frac{7}{4}$, and thus $a + r = \boxed{\frac{5}{2}}$.

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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