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Difference between revisions of "2011 AIME II Problems/Problem 5"
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==Problem==
 
==Problem==
  
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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.
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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.
  
 
==Solution==
 
==Solution==
 +
Since the sum of the first <math>2011</math> terms is <math>200</math>, and the sum of the first <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.
  
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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>.
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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>.
  
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Thus,
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==Solution 2==
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200+180+162=542
 
  
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Sum of the first 6033 is <math>\framebox[1.3\width]{542.}</math>
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Solution by e_power_pi_times_i
 +
 
 +
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>.
 +
 
 +
==Solution 3==
 +
 
 +
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 380, 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>.
 +
 
 +
==Video Solution==
 +
 
 +
https://www.youtube.com/watch?v=rpYphKOIKRs&t=186s
 +
~anellipticcurveoverq
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 +
==See also==
 +
{{AIME box|year=2011|n=II|num-b=4|num-a=6}}
 +
 
 +
[[Category:Intermediate Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 20:32, 20 January 2024

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 first $4022$ 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 380, 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 (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

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