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

 
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Problem:
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==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 <math>2011</math> terms of a [[geometric sequence]] is <math>200</math>. The sum of the first <math>4022</math> terms is <math>380</math>. Find the sum of the first <math>6033</math> terms.
  
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==Solution==
Solution: 542
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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>.
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This is decreasing from the first 2011, so the common ratio is less than one.
  
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.
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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>.
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 9/10. Following the same pattern, the sum of the third 2011 terms is (9/10)*180 = 162.
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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>.
  
Thus,
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==Solution 2==
200+180+162=542
 
  
Sum of the first 6033 is 542.
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Solution by e_power_pi_times_i
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The sum of the first <math>2011</math> terms can be written as <math>\dfrac{a_1(1-k^{2011})}{1-k}</math>, and the first <math>4022</math> terms can be written as <math>\dfrac{a_1(1-k^{4022})}{1-k}</math>. Dividing these equations, we get <math>\dfrac{1-k^{2011}}{1-k^{4022}} = \dfrac{10}{19}</math>. Noticing that <math>k^{4022}</math> is just the square of <math>k^{2011}</math>, we substitute <math>x = k^{2011}</math>, so <math>\dfrac{1}{x+1} = \dfrac{10}{19}</math>. That means that <math>k^{2011} = \dfrac{9}{10}</math>. Since the sum of the first <math>6033</math> terms can be written as <math>\dfrac{a_1(1-k^{6033})}{1-k}</math>, dividing gives <math>\dfrac{1-k^{2011}}{1-k^{6033}}</math>. Since <math>k^{6033} = \dfrac{729}{1000}</math>, plugging all the values in gives <math>\boxed{542}</math>.
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==Solution 3==
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The sum of the first 2011 terms of the sequence is expressible as <math>a_1 + a_1r + a_1r^2 + a_1r^3</math> .... until <math>a_1r^{2010}</math>. The sum of the 2011 terms following the first 2011 is expressible as <math>a_1r^{2011} + a_1r^{2012} + a_1r^{2013}</math> .... until <math>a_1r^{4021}</math>. Notice that the latter sum of terms can be expressed as <math>(r^{2011})(a_1 + a_1r + a_1r^2 + a_1r^3...a_1r^{2010})</math>. We also know that the latter sum of terms can be obtained by subtracting 200 from 180, which then means that <math>r^{2011} = 9/10</math>. The terms from 4023 to 6033 can be expressed as <math>(r^{4022})(a_1 + a_1r + a_1r^2 + a_1r^3...a_1r^{2010})</math>, which is equivalent to <math>((9/10)^2)(200) = 162</math>. Adding 380 and 162 gives the answer of <math>\boxed{542}</math>.
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==Video Solution==
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https://www.youtube.com/watch?v=rpYphKOIKRs&t=186s
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~anellipticcurveoverq
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==See also==
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{{AIME box|year=2011|n=II|num-b=4|num-a=6}}
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[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 16:41, 22 August 2020

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 $\boxed{542}$.

Solution 2

Solution by e_power_pi_times_i

The sum of the first $2011$ terms can be written as $\dfrac{a_1(1-k^{2011})}{1-k}$, and the first $4022$ terms can be written as $\dfrac{a_1(1-k^{4022})}{1-k}$. Dividing these equations, we get $\dfrac{1-k^{2011}}{1-k^{4022}} = \dfrac{10}{19}$. Noticing that $k^{4022}$ is just the square of $k^{2011}$, we substitute $x = k^{2011}$, so $\dfrac{1}{x+1} = \dfrac{10}{19}$. That means that $k^{2011} = \dfrac{9}{10}$. Since the sum of the first $6033$ terms can be written as $\dfrac{a_1(1-k^{6033})}{1-k}$, dividing gives $\dfrac{1-k^{2011}}{1-k^{6033}}$. Since $k^{6033} = \dfrac{729}{1000}$, plugging all the values in gives $\boxed{542}$.

Solution 3

The sum of the first 2011 terms of the sequence is expressible as $a_1 + a_1r + a_1r^2 + a_1r^3$ .... until $a_1r^{2010}$. The sum of the 2011 terms following the first 2011 is expressible as $a_1r^{2011} + a_1r^{2012} + a_1r^{2013}$ .... until $a_1r^{4021}$. Notice that the latter sum of terms can be expressed as $(r^{2011})(a_1 + a_1r + a_1r^2 + a_1r^3...a_1r^{2010})$. We also know that the latter sum of terms can be obtained by subtracting 200 from 180, which then means that $r^{2011} = 9/10$. The terms from 4023 to 6033 can be expressed as $(r^{4022})(a_1 + a_1r + a_1r^2 + a_1r^3...a_1r^{2010})$, which is equivalent to $((9/10)^2)(200) = 162$. Adding 380 and 162 gives the answer of $\boxed{542}$.

Video Solution

https://www.youtube.com/watch?v=rpYphKOIKRs&t=186s ~anellipticcurveoverq

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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