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

(Solution 2)
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===Solution 2===
 
===Solution 2===
After the adding of the odd numbers, the total of the sequence increases by <math>836 - 715 = 121 = 11^2</math>. Since the sum of the first <math>11</math> positive odd numbers is <math>11^2</math>, there must be <math>11</math> terms in the sequence, so the mean of the sequence is <math>\dfrac{715}{11} = 65</math>. Since the first, last, and middle terms are centered around the mean, our final answer is <math>65 \cdot 3 = \boxed{195}</math>
+
After the adding of the odd numbers, the total of the sequence increases by <math>836 - 715 = 121 = 11^2</math>. Since the sum of the first <math>n</math> positive odd numbers is <math>n^2</math>, there must be <math>11</math> terms in the sequence, so the mean of the sequence is <math>\dfrac{715}{11} = 65</math>. Since the first, last, and middle terms are centered around the mean, our final answer is <math>65 \cdot 3 = \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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:52, 20 September 2015

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.

Solutions

Solution 1

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 $\boxed{195.}$

Solution 2

After the adding of the odd numbers, the total of the sequence increases by $836 - 715 = 121 = 11^2$. Since the sum of the first $n$ positive odd numbers is $n^2$, there must be $11$ terms in the sequence, so the mean of the sequence is $\dfrac{715}{11} = 65$. Since the first, last, and middle terms are centered around the mean, our final answer is $65 \cdot 3 = \boxed{195}$

See also

2012 AIME I (ProblemsAnswer KeyResources)
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

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