Difference between revisions of "2015 AMC 10A Problems/Problem 7"
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==Solution 2== | ==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> | 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== | ==Solution 3== | ||
Minus each of the terms by 12 to make the the sequence 1,4,7,....,61. | Minus each of the terms by 12 to make the the sequence 1,4,7,....,61. | ||
61-1/3=20 20+1=21 | 61-1/3=20 20+1=21 | ||
− | <math>\boxed{\textbf{(B)}}</math>. | + | <math>\boxed{\textbf{(B)}\ 21}</math>. |
==Solution 4== | ==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 | 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 | ||
+ | <math>\boxed{\textbf{(B)}\ 21}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2015|ab=A|num-b=6|num-a=8}} | {{AMC10 box|year=2015|ab=A|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 12:29, 16 January 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 12 to make the the sequence 1,4,7,....,61. 61-1/3=20 20+1=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 .
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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