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 <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== | |
− | + | 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. | ||
− | + | 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, <math>200+180+162=542</math>, so the sum of the first <math>6033</math> terms is <math>\boxed{542}</math>. | |
− | + | ==Solution 2== | |
− | |||
− | + | 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 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>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=rpYphKOIKRs&t=186s | ||
+ | ~anellipticcurveoverq | ||
+ | |||
+ | ==See also== | ||
+ | {{AIME box|year=2011|n=II|num-b=4|num-a=6}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 16:41, 22 August 2020
Problem
The sum of the first terms of a geometric sequence is . The sum of the first terms is . Find the sum of the first 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 .
Solution 2
Solution by e_power_pi_times_i
The sum of the first terms can be written as , and the first terms can be written as . Dividing these equations, we get . Noticing that is just the square of , we substitute , so . That means that . Since the sum of the first terms can be written as , dividing gives . Since , plugging all the values in gives .
Solution 3
The sum of the first 2011 terms of the sequence is expressible as .... until . The sum of the 2011 terms following the first 2011 is expressible as .... until . Notice that the latter sum of terms can be expressed as . We also know that the latter sum of terms can be obtained by subtracting 200 from 180, which then means that . The terms from 4023 to 6033 can be expressed as , which is equivalent to . Adding 380 and 162 gives the answer of .
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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