Difference between revisions of "2012 AIME I Problems/Problem 2"

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==Solution==
 
==Solution==
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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 $715$. 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

If the sum of the original sequence is $\sum_{i=1}^{n} a_i$ then the sum of the new sequence can be expressed as $\sum_{i=1}^{n} a_i + (2i - 1) = n^2 + \sum_{i=1}^{n} a_i.$ Therefore, $836 = n^2 + 715 \rightarrow n=11.$ Now the middle term of the original sequence is simply the average of all the terms, or $\frac{715}{11} = 65,$ and the first and last terms average to this middle term, so the desired sum is simply three times the middle term, or $\boxed{195.}$

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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