2011 AIME II Problems/Problem 5

Revision as of 16:18, 13 July 2011 by Trident (talk | contribs) (Solution)

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 (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 $\frac{9}{10}$. 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 $\framebox[1.3\width]{542.}$

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