Difference between revisions of "2011 AIME II Problems/Problem 5"
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==Problem== | ==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. | + | 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== | ==Solution== | ||
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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 is less than one. | This is decreasing from the first 2011, so the common ratio is less than one. | ||
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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>. | ||
− | Thus, | + | Thus, <math>200+180+162=542</math>, so the sum of the first <math>6033</math> terms is <math>\boxed{542}</math>. |
− | <math>200+180+162=542</math> | + | |
+ | ==See also== | ||
+ | {{AIME box|year=2011|n=II|num-b=4|num-a=6}} | ||
− | + | [[Category:Intermediate Algebra Problems]] |
Revision as of 10:35, 23 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, , so the sum of the first terms is .
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 |