Difference between revisions of "2010 AMC 12A Problems/Problem 10"
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== Solution 2 == | == Solution 2 == | ||
Since all the answer choices are around <math>2010 \cdot 4 = 8040</math>, the common difference must be <math>4</math>. The first term is therefore <math>9 - 4 = 5</math>, so the <math>2010^\text{th}</math> term is <math>5 + 4 \cdot 2009 = \boxed{\textbf{(A) }8041}</math>. | Since all the answer choices are around <math>2010 \cdot 4 = 8040</math>, the common difference must be <math>4</math>. The first term is therefore <math>9 - 4 = 5</math>, so the <math>2010^\text{th}</math> term is <math>5 + 4 \cdot 2009 = \boxed{\textbf{(A) }8041}</math>. | ||
| + | |||
| + | ==Video Solution 1== | ||
| + | https://youtu.be/3gsf_XbOhhY | ||
| + | |||
| + | ~Education, the Study of Everything | ||
== See also == | == See also == | ||
Latest revision as of 19:47, 27 October 2022
Problem
The first four terms of an arithmetic sequence are 
, 
, 
, and 
. What is the 
term of this sequence?
Solution 1
and are consecutive terms, so the common difference is 
.
The common difference is 
. The first term is 
and the term is
![\[5+4(2009) = \boxed{\textbf{(A) }8041}\]](http://latex.artofproblemsolving.com/b/9/c/b9cc4cffb29eceab27e0e2bdaabebbbf4992345b.png)
Solution 2
Since all the answer choices are around 
, the common difference must be . The first term is therefore 
, so the term is 
.
Video Solution 1
~Education, the Study of Everything
See also
| 2010 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 9 |
Followed by Problem 11 |
| 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 12 Problems and Solutions | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
