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

Revision as of 16:12, 1 August 2019

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 1

$3p-q$ 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

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

Solution 2

Since all the answer choices are around $2010 \cdot 4 = 8040$, the common difference must be $4$. The first term is therefore $9 - 4 = 5$, so the $2010^\text{th}$ term is $5 + 4 \cdot 2009 = \boxed{\textbf{(A) }8041}$.

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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