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

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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, <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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==Solution 2==
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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>.
  
 
==See also==
 
==See also==

Revision as of 12:50, 1 July 2016

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 $\boxed{542}$.

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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