2009 AMC 12A 第 7 题

Question

一个等差数列的前三项分别是

2x - 3

、

5x - 11

和

3x + 1

。这个数列的第

n

项是

2009

。求

n

？

The first three terms of an arithmetic sequence are

2x - 3,

5x - 11,

and

3x + 1

respectively. The

n

th term of the sequence is

2009.

What is

n?

255

502

1004

1506

8037

答案：B

知识点：等差数列

难度评级：1250

小提示：

相邻两项的差相等：

(5x - 11) - (2x - 3)

= (3x + 1) - (5x - 11)

。

Consecutive differences are equal:

(5x - 11) - (2x - 3)

= (3x + 1) - (5x - 11)

大提示：

先求出

x

，再找到首项和公差。

Solve for

x,

then find the first term and common difference

解答：

相邻差相等给出

(5x - 11) - (2x - 3)

= (3x + 1) - (5x - 11)

，即

3x - 8 = -2x + 12

，所以

x = 4

。

前三项是

5, 9, 13

，公差为

4

。

第

n

项满足

2009 = 5 + (n - 1)\cdot 4

，所以

n - 1 = 501

，从而

n = 502

。

因此，正确答案是 B。

Equal consecutive differences give

(5x - 11) - (2x - 3)

= (3x + 1) - (5x - 11),

that is

3x - 8 = -2x + 12,

so

x = 4.

The first three terms are

5, 9, 13,

with common difference

4.

The

n

th term satisfies

2009 = 5 + (n - 1)\cdot 4,

so

n - 1 = 501

and

n = 502.

Thus, the correct answer is B.

2009 AMC 12A 第 7 题

答案：B

难度评级：1250

小提示：

大提示：

解答：

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