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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==Solution==
  
 
== 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 $715$. The first term of the sequence is increased by $1$, the second term is increased by $3$, the third term is increased by $5$, and in general, the $k$th term is increased by the $k$th odd positive integer. The terms of the new sequence add to $836$. Find the sum of the first, last, and middle terms of the original sequence.

Solution

See also

2012 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AIME Problems and Solutions
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