Difference between revisions of "2015 AMC 10A Problems/Problem 7"
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| − | ==Solution | + | ==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== | ==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 15:48, 12 February 2019
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
