Difference between revisions of "2012 AIME I Problems/Problem 2"
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− | == Problem | + | ==Problem== |
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. | ||
==Solutions== | ==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> | ||
− | + | ||
− | + | Alternatively, notice that in the original sequence, <math>11a_1 + 55d = 715</math>, from which <math>a_1 + 5d = 65</math>. Since we are tasked to find <math>a_1 + a_6 + a_{11} = 3(a_1 + 5d)</math>, the desired answer is <math>3 \cdot 65 = \boxed{195}.</math> | |
===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>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> | 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> | ||
+ | |||
+ | ===Solution 3=== | ||
+ | Proceed as in Solution 2 until it is noted that there are 11 terms in the sequence. Since the sum of the terms of the original arithmetic sequence is 715, we note that <math>\frac{2a_1 + 10d}{2} \cdot 11 = 715</math> or <math>2a_1 + 10d = 130</math> for all sets of first terms and common differences that fit the conditions given in the problem. Assume WLOG that <math>a_1 = 60</math> and <math>d = 1</math>. Then the first term of the corresponding arithmetic sequence will be <math>60</math>, the sixth (middle) term will be <math>65</math>, and the eleventh (largest) term will be <math>70</math>. Thus, our final answer is <math>60 + 65 + 70 = \boxed{195}</math>. | ||
+ | |||
+ | ~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi] | ||
+ | |||
+ | == Video Solution by Richard Rusczyk == | ||
+ | |||
+ | https://artofproblemsolving.com/videos/amc/2012aimei/298 | ||
+ | |||
+ | ~ dolphin7 | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=T8Ox412AkZc | ||
+ | ~Shreyas S | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/tKsYSBdeVuw?t=4689 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
== 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}} |
Latest revision as of 18:30, 20 January 2024
Contents
Problem
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
Alternatively, notice that in the original sequence, , from which
. Since we are tasked to find
, the desired answer is
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
Solution 3
Proceed as in Solution 2 until it is noted that there are 11 terms in the sequence. Since the sum of the terms of the original arithmetic sequence is 715, we note that or
for all sets of first terms and common differences that fit the conditions given in the problem. Assume WLOG that
and
. Then the first term of the corresponding arithmetic sequence will be
, the sixth (middle) term will be
, and the eleventh (largest) term will be
. Thus, our final answer is
.
~ cxsmi
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2012aimei/298
~ dolphin7
Video Solution
https://www.youtube.com/watch?v=T8Ox412AkZc ~Shreyas S
Video Solution by OmegaLearn
https://youtu.be/tKsYSBdeVuw?t=4689
~ pi_is_3.14
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.