Difference between revisions of "2012 AIME I Problems/Problem 2"
m |
|||
Line 3: | Line 3: | ||
==Solution== | ==Solution== | ||
+ | If the sum of the original sequence is <math>\sum_{i=1}^{n} a_i</math> then the sum of the new sequence can be expressed as <math>\sum_{i=1}^{n} a_i + (2i - 1) = n^2 + \sum_{i=1}^{n} a_i.</math> Therefore, <math>836 = n^2 + 715 \rightarrow n=11.</math> Now the middle term of the original sequence is simply the average of all the terms, or <math>\frac{715}{11} = 65,</math> and the first and last terms average to this middle term, so the desired sum is simply three times the middle term, or <math>\boxed{195.}</math> | ||
== See also == | == See also == | ||
{{AIME box|year=2012|n=I|num-b=1|num-a=3}} | {{AIME box|year=2012|n=I|num-b=1|num-a=3}} |
Revision as of 01:55, 17 March 2012
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.
Solution
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
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 |