Difference between revisions of "2012 AIME I Problems/Problem 2"
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== Problem 2 == | == Problem 2 == | ||
The terms of an arithmetic sequence add to <math>715</math>. The first term of the sequence is increased by <math>1</math>, the second term is increased by <math>3</math>, the third term is increased by <math>5</math>, and in general, the <math>k</math>th term is increased by the <math>k</math>th odd positive integer. The terms of the new sequence add to <math>836</math>. Find the sum of the first, last, and middle terms of the original sequence. | The terms of an arithmetic sequence add to <math>715</math>. The first term of the sequence is increased by <math>1</math>, the second term is increased by <math>3</math>, the third term is increased by <math>5</math>, and in general, the <math>k</math>th term is increased by the <math>k</math>th odd positive integer. The terms of the new sequence add to <math>836</math>. Find the sum of the first, last, and middle terms of the original sequence. | ||
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== 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 01:49, 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
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 |