AMC12 2002 B

AMC12 2002 B · Q9

AMC12 2002 B · Q9. It mainly tests Rational expressions, Sequences & recursion (algebra).

If $a,b,c,d$ are positive real numbers such that $a,b,c,d$ form an increasing arithmetic sequence and $a,b,d$ form a geometric sequence, then $\frac ad$ is

若 $a,b,c,d$ 是正实数，且 $a,b,c,d$ 形成递增等差数列，$a,b,d$ 形成等比数列，则 $\frac ad$ 是

(A) \frac{1}{12} \frac{1}{12}

(B) \frac{1}{6} \frac{1}{6}

(D) \frac{1}{3} \frac{1}{3}

(E) \frac{1}{2} \frac{1}{2}

Answer

Correct choice: (C)

正确答案：（C）

Solution

We can let $a=1$, $b=2$, $c=3$, and $d=4$. $\frac{a}{d}=\boxed{\frac{1}{4}} \Longrightarrow \mathrm{(C)}$ As $a, b, d$ is a geometric sequence, let $b=ka$ and $d=k^2a$ for some $k>0$. Now, $a, b, c, d$ is an arithmetic sequence. Its difference is $b-a=(k-1)a$. Thus $d=a + 3(k-1)a = (3k-2)a$. Comparing the two expressions for $d$ we get $k^2=3k-2$. The positive solution is $k=2$, and $\frac{a}{d}=\frac{a}{k^2a}=\frac{1}{k^2}=\boxed{\frac{1}{4}} \Longrightarrow \mathrm{(C)}$. Letting $n$ be the common difference of the arithmetic progression, we have $b = a + n$, $c = a + 2n$, $d = a + 3n$. We are given that $b / a$ = $d / b$, or \[\frac{a + n}{a} = \frac{a + 3n}{a + n}.\] Cross-multiplying, we get \[a^2 + 2an + n^2 = a^2 + 3an\] \[n^2 = an\] \[n = a\] So $\frac{a}{d} = \frac{a}{a + 3n} = \frac{a}{4a} = \boxed{\frac{1}{4}} \Longrightarrow \mathrm{(C)}$.

我们可以取 $a=1$、$b=2$、$c=3$、$d=4$。则 $\frac{a}{d}=\boxed{\frac{1}{4}} \Longrightarrow \mathrm{(C)}$ 由于 $a, b, d$ 是等比数列，设 $b=ka$ 且 $d=k^2a$，其中 $k>0$。又因为 $a, b, c, d$ 是等差数列，其公差为 $b-a=(k-1)a$。因此 $d=a + 3(k-1)a = (3k-2)a$。比较 $d$ 的两种表达式得 $k^2=3k-2$。其正解为 $k=2$，于是 $\frac{a}{d}=\frac{a}{k^2a}=\frac{1}{k^2}=\boxed{\frac{1}{4}} \Longrightarrow \mathrm{(C)}$。设 $n$ 为等差数列的公差，则 $b = a + n$，$c = a + 2n$，$d = a + 3n$。已知 $b / a$ = $d / b$，即 \[\frac{a + n}{a} = \frac{a + 3n}{a + n}.\] 交叉相乘得 \[a^2 + 2an + n^2 = a^2 + 3an\] \[n^2 = an\] \[n = a\] 所以 $\frac{a}{d} = \frac{a}{a + 3n} = \frac{a}{4a} = \boxed{\frac{1}{4}} \Longrightarrow \mathrm{(C)}$。

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