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2009 AMC 12A Problems/Problem 17

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Contents

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 \te...

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.

See Also

2009 AMC 12A (ProblemsResources)
Preceded by
Problem 16
Followed by
Problem 18
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