Difference between revisions of "2015 AMC 10A Problems/Problem 7"
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However, one more must be added to include the first term. So, the answer is <math>\boxed{\textbf{(B)}\ 21}</math>. | However, one more must be added to include the first term. So, the answer is <math>\boxed{\textbf{(B)}\ 21}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | <math>\dfrac{73 - 13}{3} + 1 = 21</math> <math>\boxed{\textbf{(B)}}</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}} |
Revision as of 20:50, 4 February 2015
Contents
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
.
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.