2011 AIME II Problems/Problem 5
Contents
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 terms is , and the sum of the fist terms is , the sum of the second terms is . 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 , second is , the ratio of the second terms to the first terms is . Following the same pattern, the sum of the third terms is .
Thus, , so the sum of the first terms is .
Solution 2
Solution by e_power_pi_times_i
The sum of the first terms can be written as , and the first terms can be written as . Dividing these equations, we get . Noticing that is just the square of , we substitute , so . That means that . Since the sum of the first terms can be written as , dividing gives . Since , plugging all the values in gives .
See also
2011 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.