Difference between revisions of "2012 AIME I Problems/Problem 2"
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− | Alternatively, notice that in the original sequence, <math>11a_1 + 55d = 715</math>, from which <math>a_1 + 5d = 65</math>. Since we are tasked to find <math>a_1 + a_6 + a_{11} = 3(a_1 + 5d)</math>, the desired answer is <math>3 \cdot 65 = \boxed{195. | + | Alternatively, notice that in the original sequence, <math>11a_1 + 55d = 715</math>, from which <math>a_1 + 5d = 65</math>. Since we are tasked to find <math>a_1 + a_6 + a_{11} = 3(a_1 + 5d)</math>, the desired answer is <math>3 \cdot 65 = \boxed{195}.</math> |
===Solution 2=== | ===Solution 2=== |
Revision as of 10:19, 8 September 2020
Contents
Problem 2
The terms of an arithmetic sequence add to . The first term of the sequence is increased by , the second term is increased by , the third term is increased by , and in general, the th term is increased by the th odd positive integer. The terms of the new sequence add to . Find the sum of the first, last, and middle terms of the original sequence.
Solutions
Solution 1
If the sum of the original sequence is then the sum of the new sequence can be expressed as Therefore, Now the middle term of the original sequence is simply the average of all the terms, or and the first and last terms average to this middle term, so the desired sum is simply three times the middle term, or
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Alternatively, notice that in the original sequence, , from which . Since we are tasked to find , the desired answer is
Solution 2
After the adding of the odd numbers, the total of the sequence increases by . Since the sum of the first positive odd numbers is , there must be terms in the sequence, so the mean of the sequence is . Since the first, last, and middle terms are centered around the mean, our final answer is
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2012aimei/298
~ dolphin7
Video Solution
https://www.youtube.com/watch?v=T8Ox412AkZc ~Shreyas S
See also
2012 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.