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

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

@@ Line 4: / Line 4: @@
 ==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 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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