Difference between revisions of "2015 AMC 10A Problems/Problem 7"
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==Solution 4== | ==Solution 4== | ||
− | Subtract each of the terms by 10 to make the sequence 3,6,9, | + | Subtract each of the terms by 10 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>. | <math>\boxed{\textbf{(B)}\ 21}</math>. | ||
Revision as of 18:55, 10 March 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 .
.
Solution 4
Subtract each of the terms by 10 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. .
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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