Difference between revisions of "2012 AIME I Problems/Problem 2"
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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. | ||
| − | ==Solution== | + | ==Solutions== |
| + | |||
| + | ===Solution 1=== | ||
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> | 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> | ||
| + | |||
| + | ===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> | ||
== 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 14:39, 24 August 2012
Problem 2
The terms of an arithmetic sequence add to 
. The first term of the sequence is increased by 
, the second term is increased by 
, the third term is increased by 
, and in general, the 
th term is increased by the th odd positive integer. The terms of the new sequence add to 
. 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 
then the sum of the new sequence can be expressed as 
Therefore, 
Now the middle term of the original sequence is simply the average of all the terms, or 
and the first and last terms average to this middle term, so the desired sum is simply three times the middle term, or 
Solution 2
After the adding of the odd numbers, the total of the sequence increases by 
. Since the sum of the first 
positive odd numbers is 
, there must be terms in the sequence, so the mean of the sequence is 
. Since the first, last, and middle terms are centered around the mean, our final answer is 
See also
| 2012 AIME I (Problems • Answer Key • Resources) | ||
| 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 | ||
