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

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==Problem==
 
== 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.

Revision as of 15:42, 9 August 2018

Problem

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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