2012 AIME I Problems/Problem 2

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 $\boxed{195.}$

2012 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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2012 AIME I Problems/Problem 2

Problem 2

Solution

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