Difference between revisions of "2015 AMC 10A Problems/Problem 7"
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==Solution 3== | ==Solution 3== | ||
− | Minus each of the terms by 12 to make the the sequence <math>1 , 4 , 7,..., 61</math>. | + | 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> |
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− | <math>61-1/3=20, 20 + 1 = 21</math> | ||
<math>\boxed{\textbf{(B)}\ 21}</math>. | <math>\boxed{\textbf{(B)}\ 21}</math>. |
Revision as of 21:20, 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 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.
.
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.