AMC12 2010 A

AMC12 2010 A · Q10

AMC12 2010 A · Q10. It mainly tests Systems of equations, Sequences & recursion (algebra).

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?

一个等差数列的前四项是 $p$、$9$、$3p-q$ 和 $3p+q$。该数列的第 $2010^\text{th}$ 项是多少？

(A) 8041 8041

(B) 8043 8043

(D) 8047 8047

(E) 8049 8049

Answer

Correct choice: (A)

正确答案：（A）

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

$3p-q$ 和 $3p+q$ 是相邻两项，所以公差为 $(3p+q)-(3p-q)=2q$。 \begin{align*}p+2q &= 9\\ 9+2q &= 3p-q\\ q&=2\\ p&=5\end{align*} 公差为 $4$。首项为 $5$，第 $2010^\text{th}$ 项为 \[5+4(2009) = \boxed{\textbf{(A) }8041}\]

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