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

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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 | + | 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 18:36, 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 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,

Sum of the first is