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

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==Problem==
 
==Problem==
  
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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>
  
 +
==Solution 1==
  
 +
<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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==Solution==
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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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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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However, the number 0 must also be included, adding another multiple. So, the answer is <math>\boxed{\textbf{(B)}\ 21}</math>.
  
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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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==Solution 2==
 +
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>
 +
<math>\boxed{\textbf{(B)}\ 21}</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>.
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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>\frac{61-1}{3}=20, 20 + 1 = 21</math>
 +
 
 +
<math>\boxed{\textbf{(B)}\ 21}</math>.
 +
 
 +
==Solution 4==
 +
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.
 +
<math>\boxed{\textbf{(B)}\ 21}</math>.
 +
 
 +
==Video Solution (CREATIVE THINKING)==
 +
https://youtu.be/i7ItueJ6K8E
 +
 
 +
~Education, the Study of Everything
 +
 
 +
 
 +
 
 +
 
 +
==Video Solution==
 +
https://youtu.be/fcWPfgKeCmA
 +
 
 +
~savannahsolver
 +
 
 +
==See Also==
 +
{{AMC10 box|year=2015|ab=A|num-b=6|num-a=8}}
 +
{{MAA Notice}}

Latest revision as of 22:05, 26 June 2023

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 1

$73-13 = 60$, 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, the number 0 must also be included, adding another multiple. 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 $1 , 4 , 7,..., 61$. $\frac{61-1}{3}=20, 20 + 1 = 21$

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

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 (CREATIVE THINKING)

https://youtu.be/i7ItueJ6K8E

~Education, the Study of Everything



Video Solution

https://youtu.be/fcWPfgKeCmA

~savannahsolver

See Also

2015 AMC 10A (Problems • Answer Key • Resources)
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

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

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