2009 AMC 12A Problems/Problem 17

Problem

Let $a + ar_1 + ar_1^2 + ar_1^3 + \cdots$ and $a + ar_2 + ar_2^2 + ar_2^3 + \cdots$ be two different infinite geometric series of positive numbers with the same first term. The sum of the first series is $r_1$ , and the sum of the second series is $r_2$ . What is $r_1 + r_2$ ?

$\textbf{(A)}\ 0\qquad \textbf{(B)}\ \frac {1}{2}\qquad \textbf{(C)}\ 1\qquad \textbf{(D)}\ \frac {1 + \sqrt {5}}{2}\qquad \textbf{(E)}\ 2$

Solution

Using the formula for the sum of a geometric series we get that the sums of the given two sequences are $\frac a{1-r_1}$ and $\frac a{1-r_2}$ .

Hence we have $\frac a{1-r_1} = r_1$ and $\frac a{1-r_2} = r_2$ . This can be rewritten as $r_1(1-r_1) = r_2(1-r_2) = a$ .

As we are given that $r_1$ and $r_2$ are distinct, these must be precisely the two roots of the equation $x^2 - x + a = 0$ .

Using Vieta's formulas we get that the sum of these two roots is $\boxed{1}$ .

Alternate Solution

Using the formula for the sum of a geometric series we get that the sums of the given two sequences are $\frac a{1-r_1}$ and $\frac a{1-r_2}$ .

Hence we have $\frac a{1-r_1} = r_1$ and $\frac a{1-r_2} = r_2$ . This can be rewritten as $r_1(1-r_1) = r_2(1-r_2) = a$ .

Which can be further rewritten as $r_1-r_1^2 = r_2-r_2^2$ . Rearranging the equation we get $r_1-r_2 = r_1^2-r_2^2$ . Expressing this as a difference of squares we get $r_1-r_2 = (r_1-r_2)(r_1+r_2)$ .

Dividing by like terms we finally get $r_1+r_2 = \boxed{1}$ as desired.

2009 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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2009 AMC 12A Problems/Problem 17

Contents

Problem

Solution

Alternate Solution

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