# 2010 AMC 12A Problems/Problem 10

## 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 $5 + 4 \cdot 2009 = \boxed{\textbf{(A) }8041}$.

## Video Solution 1

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