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

(Solution 1)
Line 1: Line 1:
== Problem 2 ==
+
==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==
+
==Solution 1==
 
 
===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>
  
Line 11: Line 9:
 
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>
 
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>
  
=== Video Solution by Richard Rusczyk ===
+
== Video Solution by Richard Rusczyk ==
  
 
https://artofproblemsolving.com/videos/amc/2012aimei/298
 
https://artofproblemsolving.com/videos/amc/2012aimei/298

Revision as of 20:17, 24 January 2021

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.

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

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2012aimei/298

~ dolphin7

Video Solution

https://www.youtube.com/watch?v=T8Ox412AkZc ~Shreyas S

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png