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.
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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==
 
 
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,  
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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>
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==See also==
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{{AIME box|year=2011|n=II|num-b=4|num-a=6}}
  
Sum of the first <math>6033</math> is <math>\framebox[1.3\width]{542.}</math>
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[[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 $2011$ 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 $\boxed{542}$.

See also

2011 AIME II (ProblemsAnswer KeyResources)
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
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