Difference between revisions of "2015 AMC 10A Problems/Problem 7"
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==Problem== | ==Problem== | ||
− | How many terms are in the arithmetic sequence 13, 16, 19, | + | 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{( | + | <math> \textbf{(A)}\ 20 \qquad\textbf{(B)} \ 21 \qquad\textbf{(C)} \ 24 \qquad\textbf{(D)} \ 60 \qquad\textbf{(E)} \ 61 </math> |
+ | ==Solution== | ||
+ | |||
+ | <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>. | ||
+ | |||
+ | 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>. | ||
+ | However, one more must be added to include the first term. So, the answer is <math>\boxed{\textbf{(B)}\ 21}</math>. | ||
− | ==Solution== | + | ==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>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | 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> | ||
+ | |||
+ | <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== | |
+ | 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 15:34, 7 July 2020
Problem
How many terms are in the arithmetic sequence , , , , , ?
Solution
, so the amount of terms in the sequence , , , , , is the same as in the sequence , , , , , .
In this sequence, the terms are the multiples of going up to , and there are multiples of in .
However, one more must be added to include the first term. So, the answer is .
Solution 2
Using the formula for arithmetic sequence's nth term, we see that .
Solution 3
Minus each of the terms by to make the the sequence .
.
Solution 4
Subtract each of the terms by to make the sequence . Then divide the each term in the sequence by to get . Now it is clear to see that there are terms in the sequence. .
Video Solution
~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 |
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