2008 AMC 12A 第 16 题

Question

数

\log(a^3 b^7)

、

\log(a^5 b^{12})

、

\log(a^8 b^{15})

是一个等差数列的前三项，且该数列第

12

项为

\log(b^n)

。

n

是多少？

The numbers

\log(a^3 b^7),

\log(a^5 b^{12}),

and

\log(a^8 b^{15})

are the first three terms of an arithmetic sequence, and the

12

th term of the sequence is

\log(b^n).

What is

n?

40

56

76

112

143

答案：D

知识点：等差数列对数

难度评级：1800

小提示：

把每一项写成

p \log a + q \log b

，并令相邻差相等。

Write each term as

p \log a + q \log b

and set the consecutive differences equal

大提示：

相邻差相等迫使

\log a = 2 \log b

，从而每项都化为

\log b

的倍数。

Equal differences force

\log a = 2 \log b,

reducing every term to a multiple of

\log b

解答：

三项分别为

3\log a + 7\log b

、

5\log a + 12\log b

和

8\log a + 15\log b

。令相邻两项之差相等，得

\begin{aligned} &2\log a + 5\log b \\ &= 3\log a + 3\log b \end{aligned}\text{，}

所以

\log a = 2\log b

。

于是第一项为

(3 \cdot 2 + 7)\log b = 13\log b

，公差为

(2 \cdot 2 + 5)\log b = 9\log b

。

第

12

项为

\begin{aligned} (13 + 11 \cdot 9)\log b &= 112\log b \\ &= \log(b^{112}) \end{aligned}\text{，}

所以

n = 112

。

所以正确答案是 D。

The three terms are

3\log a + 7\log b,

5\log a + 12\log b,

and

8\log a + 15\log b.

Setting the two consecutive differences equal,

\begin{aligned} &2\log a + 5\log b \\ &= 3\log a + 3\log b, \end{aligned}

so

\log a = 2\log b.

The first term is then

(3 \cdot 2 + 7)\log b = 13\log b,

and the common difference is

(2 \cdot 2 + 5)\log b = 9\log b.

The

12

th term is

\begin{aligned} (13 + 11 \cdot 9)\log b &= 112\log b \\ &= \log(b^{112}), \end{aligned}

so

n = 112.

Thus, D is the correct answer.

2008 AMC 12A 第 16 题

答案：D

难度评级：1800

小提示：

大提示：

解答：

其他年份的第 16 题