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Difference between revisions of "2012 AIME I Problems/Problem 2"
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== Problem 2 ==
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==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==
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===Solution 1===
 
===Solution 1===
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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>
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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>
  
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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>
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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>
  
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=== Video Solution by Richard Rusczyk ===
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===Solution 3===
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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>.
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~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
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==Solution 4 (Similar to the above solutions) ==
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Let the sequence be <math>a,a+d,a+2d,..., a+(k-1)d</math>. Since there are <math>k</math> terms, we have an equation <math>k^2 = 836-715 = 121</math>. Solving, we get <math>k</math> as <math>11</math>. Replacing <math>k</math> in our sequence with <math>11</math>, we get <math>a,a+d,a+2d,..., a+10d</math>. The sum of this sequence is equal to <math>715</math>, or <math>\frac{(2a+10d)(11)}{2} = 715</math>, and by simplifying we get <math>a+5d=65</math>. We are asked for the first, middle, and end terms, which are <math>a</math>,<math>a+5d</math>, and <math>a+10d</math> respectively. Their sum is <math>3a+15</math>, or <math>3 \cdot (a+5d)</math>. Our desired answer is <math>65 \cdot 3 = \boxed{195}</math>.
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~Irfans123
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== Video Solution by Richard Rusczyk ==
  
 
https://artofproblemsolving.com/videos/amc/2012aimei/298
 
https://artofproblemsolving.com/videos/amc/2012aimei/298
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https://www.youtube.com/watch?v=T8Ox412AkZc
 
https://www.youtube.com/watch?v=T8Ox412AkZc
 
~Shreyas S
 
~Shreyas S
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== Video Solution by OmegaLearn ==
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https://youtu.be/tKsYSBdeVuw?t=4689
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~ 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 20:59, 26 February 2025

Problem

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 $\boxed{195}.$


Alternatively, notice that in the original sequence, $11a_1 + 55d = 715$, from which $a_1 + 5d = 65$. Since we are tasked to find $a_1 + a_6 + a_{11} = 3(a_1 + 5d)$, the desired answer is $3 \cdot 65 = \boxed{195}.$

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 $65 \cdot 3 = \boxed{195}$

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 $\frac{2a_1 + 10d}{2} \cdot 11 = 715$ or $2a_1 + 10d = 130$ for all sets of first terms and common differences that fit the conditions given in the problem. Assume WLOG that $a_1 = 60$ and $d = 1$. Then the first term of the corresponding arithmetic sequence will be $60$, the sixth (middle) term will be $65$, and the eleventh (largest) term will be $70$. Thus, our final answer is $60 + 65 + 70 = \boxed{195}$.

~ cxsmi

Solution 4 (Similar to the above solutions)

Let the sequence be $a,a+d,a+2d,..., a+(k-1)d$. Since there are $k$ terms, we have an equation $k^2 = 836-715 = 121$. Solving, we get $k$ as $11$. Replacing $k$ in our sequence with $11$, we get $a,a+d,a+2d,..., a+10d$. The sum of this sequence is equal to $715$, or $\frac{(2a+10d)(11)}{2} = 715$, and by simplifying we get $a+5d=65$. We are asked for the first, middle, and end terms, which are $a$,$a+5d$, and $a+10d$ respectively. Their sum is $3a+15$, or $3 \cdot (a+5d)$. Our desired answer is $65 \cdot 3 = \boxed{195}$.

~Irfans123

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

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