Difference between revisions of "2015 AMC 10A Problems/Problem 7"

(Problem and Solution 7)
 
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==Problem==
 
==Problem==
  
How many terms are in the arithmetic sequence 13, 16, 19,..., 70, 73?
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How many terms are in the arithmetic sequence <math>13</math>, <math>16</math>, <math>19</math>, <math>\dotsc</math>, <math>70</math>, <math>73</math>?
  
<math> \textbf{(a)}\ 20 \qquad\textbf{(B)} \ 21 \qquad\textbf{(C)} \ 24 \qquad\textbf{(D)} \ 60 \qquad\textbf{(E)} \ 61 </math>
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<math> \textbf{(A)}\ 20 \qquad\textbf{(B)} \ 21 \qquad\textbf{(C)} \ 24 \qquad\textbf{(D)} \ 60 \qquad\textbf{(E)} \ 61 </math>
  
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==Solution==
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<math>73-13 = 60</math>, so the amount of terms in the sequence <math>13</math>, <math>16</math>, <math>19</math>, <math>\dotsc</math>, <math>70</math>, <math>73</math> is the same as in the sequence <math>0</math>, <math>3</math>, <math>6</math>, <math>\dotsc</math>, <math>57</math>, <math>60</math>.
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In this sequence, the terms are the multiples of <math>3</math> going up to <math>60</math>, and there are <math>20</math> multiples of <math>3</math> in <math>60</math>.
  
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However, one more must be added to include the first term. So, the answer is <math>\boxed{\textbf{(B)}\ 21}</math>.
  
==Solution==
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==Solution 2==
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Using the formula for arithmetic sequence's nth term, we see that <math>a + (n-1)d \Longrightarrow13 + (n-1)3 =73, \Longrightarrow n = 21</math>
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<math>\boxed{\textbf{(B)}\ 21}</math>.
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==Solution 3==
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Minus each of the terms by <math>12</math> to make the the sequence <math>1 , 4 , 7,..., 61</math>. <math>61-1/3=20, 20 + 1 = 21</math>
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<math>\boxed{\textbf{(B)}\ 21}</math>.
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==Solution 4==
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Subtract each of the terms by <math>10</math> to make the sequence <math>3 , 6 , 9,..., 60, 63</math>. Then divide the each term in the sequence by <math>3</math> to get <math>1, 2, 3,..., 20, 21</math>. Now it is clear to see that there are <math>21</math> terms in the sequence.
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<math>\boxed{\textbf{(B)}\ 21}</math>.
  
73-13 is 60, so the amount of terms in the sequence 13, 16, 19,..., 70, 73 is the same as in the sequence 0, 3, 6,..., 57, 60.  
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==Video Solution==
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https://youtu.be/fcWPfgKeCmA
  
In this sequence, the terms are the multiples of 3 going up to 60, and there are 20 multiples of 3 in 60.
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~savannahsolver
  
However, one more must be added to include the first term. So, the answer is <math>\boxed{\textbf{(B) }21}</math>.
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==See Also==
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{{AMC10 box|year=2015|ab=A|num-b=6|num-a=8}}
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{{MAA Notice}}

Latest revision as of 15:34, 7 July 2020

Problem

How many terms are in the arithmetic sequence $13$, $16$, $19$, $\dotsc$, $70$, $73$?

$\textbf{(A)}\ 20 \qquad\textbf{(B)} \ 21 \qquad\textbf{(C)} \ 24 \qquad\textbf{(D)} \ 60 \qquad\textbf{(E)} \ 61$

Solution

$73-13 = 60$, so the amount of terms in the sequence $13$, $16$, $19$, $\dotsc$, $70$, $73$ is the same as in the sequence $0$, $3$, $6$, $\dotsc$, $57$, $60$.

In this sequence, the terms are the multiples of $3$ going up to $60$, and there are $20$ multiples of $3$ in $60$.

However, one more must be added to include the first term. So, the answer is $\boxed{\textbf{(B)}\ 21}$.

Solution 2

Using the formula for arithmetic sequence's nth term, we see that $a + (n-1)d \Longrightarrow13 + (n-1)3 =73, \Longrightarrow n = 21$ $\boxed{\textbf{(B)}\ 21}$.

Solution 3

Minus each of the terms by $12$ to make the the sequence $1 , 4 , 7,..., 61$. $61-1/3=20, 20 + 1 = 21$

$\boxed{\textbf{(B)}\ 21}$.

Solution 4

Subtract each of the terms by $10$ to make the sequence $3 , 6 , 9,..., 60, 63$. Then divide the each term in the sequence by $3$ to get $1, 2, 3,..., 20, 21$. Now it is clear to see that there are $21$ terms in the sequence. $\boxed{\textbf{(B)}\ 21}$.

Video Solution

https://youtu.be/fcWPfgKeCmA

~savannahsolver

See Also

2015 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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