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

Revision as of 22:48, 4 July 2013

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, $200+180+162=542$ , so the sum of the first $6033$ terms is $\boxed{542}$ .

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

Revision as of 22:48, 4 July 2013

Problem

Solution

See also