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2009 AMC 12A Problems/Problem 17

Revision as of 20:43, 3 July 2013 by Nathan wailes (talk | contribs) (→‎See Also)

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.



Note: It is necessary to check that $r_1-r_2\ne 0$, as you cannot divide by zero. As the problem states that the series are different, $r_1 \ne r_2$, and so there is no division by zero error.

See Also

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