# 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 $\boxed{1}$.

## Solution 2

We basically have two infinite geometric series whose sum is equivalent to the common ratio. Let us have a geometric series: $b, br, br^2.....$.

The sum is: $\frac{b}{1-r} = r.$ Thus, $b = r-r^2$ and by Vieta's, the sum of the two possible values of $r$ ( $r_1$ and $r_2$) is $1$.

~conantwiz2023

## 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.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 