AMC12 2008 A

AMC12 2008 A · Q16

AMC12 2008 A · Q16. It mainly tests Exponents & radicals, Sequences & recursion (algebra).

The numbers $\log(a^3b^7)$, $\log(a^5b^{12})$, and $\log(a^8b^{15})$ are the first three terms of an arithmetic sequence, and the $12^\text{th}$ term of the sequence is $\log{b^n}$. What is $n$?

数 $\log(a^3b^7)$，$\log(a^5b^{12})$ 和 $\log(a^8b^{15})$ 是等差数列的前三项，该数列的第 $12^\text{th}$ 项是 $\log{b^n}$。$n$ 是多少？

(A) 40 40

(B) 56 56

(D) 112 112

(E) 143 143

Answer

Correct choice: (D)

正确答案：（D）

Solution

Let $A = \log(a)$ and $B = \log(b)$. The first three terms of the arithmetic sequence are $3A + 7B$, $5A + 12B$, and $8A + 15B$, and the $12^\text{th}$ term is $nB$. Thus, $2(5A + 12B) = (3A + 7B) + (8A + 15B) \Rightarrow A = 2B$. Since the first three terms in the sequence are $13B$, $22B$, and $31B$, the $k$th term is $(9k + 4)B$. Thus the $12^\text{th}$ term is $(9\cdot12 + 4)B = 112B = nB \Rightarrow n = 112\Rightarrow \boxed{D}$.

设 $A = \log(a)$，$B = \log(b)$。等差数列的前三项分别为 $3A + 7B$，$5A + 12B$，$8A + 15B$，第 $12^\text{th}$ 项为 $nB$。因此，$2(5A + 12B) = (3A + 7B) + (8A + 15B) \Rightarrow A = 2B$。由于前三项为 $13B$，$22B$，$31B$，第 $k$ 项为 $(9k + 4)B$。所以第 $12^\text{th}$ 项为 $(9\cdot12 + 4)B = 112B = nB \Rightarrow n = 112\Rightarrow \boxed{D}$。

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