# Difference between revisions of "2011 AIME II Problems/Problem 5"

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− | Solution: | + | Solution: 542 |

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. | 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 (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 (or whatever the term for what you multiply it by is) 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 9/10. Following the same pattern, the sum of the third 2011 terms is (9/10)*180 = 162. | |

− | Because it is a geometric sequence and the first 2011 terms is 200, second 2011 is 180, | ||

− | |||

− | Sum of the first 6033 is | + | Thus, |

+ | 200+180+162=542 | ||

+ | |||

+ | Sum of the first 6033 is 542. |

## Revision as of 00:17, 31 March 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: 542

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 (or whatever the term for what you multiply it by is) 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 9/10. Following the same pattern, the sum of the third 2011 terms is (9/10)*180 = 162.

Thus, 200+180+162=542

Sum of the first 6033 is 542.