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 "2011 AIME II Problems/Problem 5"

(→‎Solution)
(→‎Solution)
Line 6: Line 6:
  
 
Since the sum of the first <math>2011</math> terms is <math>200</math>, and the sum of the fist <math>4022</math> terms is <math>380</math>, the sum of the second <math>2011</math> terms is <math>180</math>.
 
Since the sum of the first <math>2011</math> terms is <math>200</math>, and the sum of the fist <math>4022</math> terms is <math>380</math>, the sum of the second <math>2011</math> terms is <math>180</math>.
−
This is decreasing from the first 2011, so the common ratio (or whatever the term for what you multiply it by is) is less than one.
+
This is decreasing from the first 2011, so the common ratio is less than one.
  
 
Because it is a geometric sequence and the sum of the first 2011 terms is <math>200</math>, second <math>2011</math> is <math>180</math>, the ratio of the second <math>2011</math> terms to the first <math>2011</math> terms is <math>\frac{9}{10}</math>. Following the same pattern, the sum of the third <math>2011</math> terms is <math>\frac{9}{10}*180 = 162</math>.
 
Because it is a geometric sequence and the sum of the first 2011 terms is <math>200</math>, second <math>2011</math> is <math>180</math>, the ratio of the second <math>2011</math> terms to the first <math>2011</math> terms is <math>\frac{9}{10}</math>. Following the same pattern, the sum of the third <math>2011</math> terms is <math>\frac{9}{10}*180 = 162</math>.

Revision as of 17:36, 16 August 2011

Problem

The sum of the first 2011 terms of a geometric sequence is 200. The sum of the first 4022 terms is 380. Find the sum of the first 6033 terms.

Solution

Since the sum of the first $2011$ terms is $200$, and the sum of the fist $4022$ terms is $380$, the sum of the second $2011$ terms is $180$. This is decreasing from the first 2011, so the common ratio is less than one.

Because it is a geometric sequence and the sum of the first 2011 terms is $200$, second $2011$ is $180$, the ratio of the second $2011$ terms to the first $2011$ terms is $\frac{9}{10}$. Following the same pattern, the sum of the third $2011$ terms is $\frac{9}{10}*180 = 162$.

Thus, $200+180+162=542$

Sum of the first $6033$ is $\framebox[1.3\width]{542.}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_II_Problems/Problem_5&oldid=41537"