Difference between revisions of "2009 AMC 12A Problems/Problem 17"

Latest revision as of 09:22, 6 October 2023

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

Using the previous solution we reach the equality $r_1(1-r_1) = r_2(1-r_2)$ .

Obviously, since $r_1 \neq r_2$ , then $r_1 = 1 - r_2$ so $r_1 + r_2 = 1$ .

-Vignesh Peddi

Solution 3

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.

@@ Line 18: / Line 18: @@
 == Solution 2 ==
-We basically have two infinite geometric series whose sum is equivalent to the common ratio. Let us have a geometric series: <math>b, br, br^2.....</math>.
+Using the previous solution we reach the equality <math>r_1(1-r_1) = r_2(1-r_2)</math>.
-The sum is: <math>\frac{b}{1-r} = r.</math> Thus, <math>b = r-r^2</math> and by Vieta's, the sum of the two possible values of <math>r</math> (<math>r_1</math> and <math>r_2</math>) is <math>1</math>.
+Obviously, since <math>r_1 \neq r_2</math>, then <math>r_1 = 1 - r_2</math> so <math>r_1 + r_2 = 1</math>.
+-Vignesh Peddi
-~conantwiz2023
+== Solution 3 ==
+We basically have two infinite geometric series whose sum is equivalent to the common ratio. Let us have a geometric series: <math>b, br, br^2.....</math>.
-== Solution 3 ==
+The sum is: <math>\frac{b}{1-r} = r.</math> Thus, <math>b = r-r^2</math> and by Vieta's, the sum of the two possible values of <math>r</math> (<math>r_1</math> and <math>r_2</math>) is <math>1</math>.
-Using the previous solutions we reach the equality <math>r_1(1-r_1) = r_2(1-r_2)</math>
-Obviously since <math>r_1 != r_2</math>, then <math>r_1 = 1 - r_2</math> so <math>r_1 + r_2 = 1</math>.
--Vignesh Peddi
+~conantwiz2023
 == Alternate Solution ==

2009 AMC 12A (Problems • Answer Key • Resources)
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