Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(New page: == Problem == Let <math>a + ar_1 + ar_1^2 + ar_1^3 + \cdots</math> and <math>a + ar_2 + ar_2^2 + ar_2^3 + \cdots</math> be two different infinite geometric series of positive numbers with ...)
 
(→‎Solution 2)
 
(13 intermediate revisions by 6 users not shown)
Line 15: Line 15:
 
Using [[Vieta's formulas]] we get that the sum of these two roots is <math>\boxed{1}</math>.
 
Using [[Vieta's formulas]] we get that the sum of these two roots is <math>\boxed{1}</math>.
  
 +
 +
== Solution 2 ==
 +
 +
Using the previous solution we reach the equality <math>r_1(1-r_1) = r_2(1-r_2)</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
 +
 +
== 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>.
 +
 +
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>.
 +
 +
 +
 +
~conantwiz2023
 +
 +
== Alternate Solution ==
 +
 +
 +
Using the formula for the sum of a geometric series we get that the sums of the given two sequences are <math>\frac a{1-r_1}</math> and <math>\frac a{1-r_2}</math>.
 +
 +
Hence we have <math>\frac a{1-r_1} = r_1</math> and <math>\frac a{1-r_2} = r_2</math>.
 +
This can be rewritten as <math>r_1(1-r_1) = r_2(1-r_2) = a</math>.
 +
 +
Which can be further rewritten as <math>r_1-r_1^2 = r_2-r_2^2</math>.
 +
Rearranging the equation we get <math>r_1-r_2 = r_1^2-r_2^2</math>.
 +
Expressing this as a difference of squares we get <math> r_1-r_2 = (r_1-r_2)(r_1+r_2)</math>.
 +
 +
Dividing by like terms we finally get <math>r_1+r_2 = \boxed{1}</math> as desired.
 +
 +
 +
 +
 +
Note: It is necessary to check that <math>r_1-r_2\ne 0</math>, as you cannot divide by zero. As the problem states that the series are different, <math>r_1 \ne r_2</math>, and so there is no division by zero error.
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC12 box|year=2009|ab=A|num-b=16|num-a=18}}
 
{{AMC12 box|year=2009|ab=A|num-b=16|num-a=18}}
 +
{{MAA Notice}}

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.

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_17&oldid=199198"