Difference between revisions of "2010 AMC 12A Problems/Problem 10"
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<math>\textbf{(A)}\ 8041 \qquad \textbf{(B)}\ 8043 \qquad \textbf{(C)}\ 8045 \qquad \textbf{(D)}\ 8047 \qquad \textbf{(E)}\ 8049</math> | <math>\textbf{(A)}\ 8041 \qquad \textbf{(B)}\ 8043 \qquad \textbf{(C)}\ 8045 \qquad \textbf{(D)}\ 8047 \qquad \textbf{(E)}\ 8049</math> | ||
− | == Solution == | + | == Solution 1 == |
<math>3p-q</math> and <math>3p+q</math> are consecutive terms, so the common difference is <math>(3p+q)-(3p-q) = 2q</math>. | <math>3p-q</math> and <math>3p+q</math> are consecutive terms, so the common difference is <math>(3p+q)-(3p-q) = 2q</math>. | ||
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The common difference is <math>4</math>. The first term is <math>5</math> and the <math>2010^\text{th}</math> term is | The common difference is <math>4</math>. The first term is <math>5</math> and the <math>2010^\text{th}</math> term is | ||
− | <cmath>5+4(2009) = \boxed{8041\ \textbf{(A)}}</ | + | <cmath>5+4(2009) = \boxed{\textbf{(A) }8041}</cmath> |
+ | |||
+ | == 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>. | ||
+ | |||
+ | ==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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.