Difference between revisions of "2015 AMC 10A Problems/Problem 7"
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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. | 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>. | <math>\boxed{\textbf{(B)}\ 21}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/fcWPfgKeCmA | ||
+ | |||
+ | ~savannahsolver | ||
==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}} |
Revision as of 14:34, 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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