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2010 AMC 12A Problems/Problem 10: Difference between revisions

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Created page with '== Problem 10 == The first four terms of an arithmetic sequence are <math>p</math>, <math>9</math>, <math>3p-q</math>, and <math>3p+q</math>. What is the <math>2010^\text{th}</ma…'
 
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== Problem 10 ==
== Problem ==
The first four terms of an arithmetic sequence are <math>p</math>, <math>9</math>, <math>3p-q</math>, and <math>3p+q</math>. What is the <math>2010^\text{th}</math> term of this sequence?
The first four terms of an arithmetic sequence are <math>p</math>, <math>9</math>, <math>3p-q</math>, and <math>3p+q</math>. What is the <math>2010^\text{th}</math> term of this sequence?


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<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>.


<cmath>\begin{align*}&p+2q = 9\\
&9+2q = 3p-q\\
&q=2\\
&p=5\end{align*}</cmath>


<math>p+2q = 9</math>


<math>9+2q = 3p-q</math>
The common difference is <math>4</math>. The first term is <math>5</math> and the <math>2010^\text{th}</math> term is
 
<math>q=2</math>
 
<math>p=5</math>


<cmath>5+4(2009) = \boxed{8041\ \textbf{(A)}}</cmath>


The common difference is <math>4</math>. The first term is <math>5</math> and the <math>2010^\text{th}</math> term is
== See also ==
{{AMC12 box|year=2010|num-b=9|num-a=11|ab=A}}


<math>5+4(2009) = \boxed{8041\ \textbf{(A)}}</math>
[[Category:Introductory Algebra Problems]]

Revision as of 22:16, 25 February 2010

Problem

The first four terms of an arithmetic sequence are $p$, $9$, $3p-q$, and $3p+q$. What is the $2010^\text{th}$ term of this sequence?

$\textbf{(A)}\ 8041 \qquad \textbf{(B)}\ 8043 \qquad \textbf{(C)}\ 8045 \qquad \textbf{(D)}\ 8047 \qquad \textbf{(E)}\ 8049$

Solution

$3p-q$ and $3p+q$ are consecutive terms, so the common difference is $(3p+q)-(3p-q) = 2q$.

\begin{align*}&p+2q = 9\\ &9+2q = 3p-q\\ &q=2\\ &p=5\end{align*}


The common difference is $4$. The first term is $5$ and the $2010^\text{th}$ term is

\[5+4(2009) = \boxed{8041\ \textbf{(A)}}\]

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
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