Question

一个等差数列的前四项为 $p$ 、 $9$ 、 $3p-q$ 和 $3p+q$ 。这个数列的第 $2010$ 项是多少？

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

$8041$

$8043$

$8045$

$8047$

$8049$

公差等于 $(3p+q)-(3p-q)=2q$ 。

The common difference equals $(3p+q)-(3p-q)=2q$

同时 $9-p=2q$ ，且 $(3p-q)-9=2q$ ；解出 $p$ 和 $q$ 。

Also $9-p=2q$ and $(3p-q)-9=2q;$ solve for $p$ and $q$

相邻两项之差都是公差 $d$ ，且 $d=(3p+q)-(3p-q)=2q$ 。

由前两项可得 $9-p=d=2q$ ，由第二、三项可得 $(3p-q)-9=d=2q$ 。解得 $p=5$ ， $q=2$ ， $d=4$ 。

因此第 $2010$ 项为 $\begin{aligned} p+2009d &= 5+2009\cdot4 \\ &= 8041 \end{aligned}\text{。}$

所以正确答案是 A。

Consecutive terms differ by a common difference $d.$ From the last two terms, $d=(3p+q)-(3p-q)=2q.$

From the first two terms, $9-p=d=2q,$ and from the second and third, $(3p-q)-9=d=2q.$ Solving this system gives $p=5,$ $q=2,$ and $d=4.$

The $2010$ th term is $\begin{aligned} p+2009d &= 5+2009\cdot4 \\ &= 8041. \end{aligned}$

Thus, A is the correct answer.