AMC12 2003 B

AMC12 2003 B · Q6

AMC12 2003 B · Q6. It mainly tests Exponents & radicals, Sequences & recursion (algebra).

The second and fourth terms of a geometric sequence are $2$ and $6$. Which of the following is a possible first term?

一个等比数列的第二项和第四项分别是 $2$ 和 $6$。以下哪一项可能是首项？

(A) -\sqrt{3} -\sqrt{3}

(B) -\frac{2\sqrt{3}}{3} -\frac{2\sqrt{3}}{3}

(D) \sqrt{3} \sqrt{3}

(E) 3 3

Answer

Correct choice: (B)

正确答案：（B）

Solution

Let the first term be $a$ and the common ratio be $r$. Therefore, \[ar=2\ \ (1) \qquad \text{and} \qquad ar^3=6\ \ (2)\] Dividing $(2)$ by $(1)$ eliminates the $a$, yielding $r^2=3$, so $r=\pm\sqrt{3}$. Now, since $ar=2$, $a=\frac{2}{r}$, so $a=\frac{2}{\pm\sqrt{3}}=\pm\frac{2\sqrt{3}}{3}$. We therefore see that $\boxed{\textbf{(B)}\ -\frac{2\sqrt{3}}{3}}$ is a possible first term.

设首项为 $a$，公比为 $r$。则 \[ar=2\ \ (1) \qquad \text{且} \qquad ar^3=6\ \ (2)\] 用 $(2)$ 除以 $(1)$ 消去 $a$，得 $r^2=3$，所以 $r=\pm\sqrt{3}$。又因为 $ar=2$，所以 $a=\frac{2}{r}$，从而 $a=\frac{2}{\pm\sqrt{3}}=\pm\frac{2\sqrt{3}}{3}$。因此 $\boxed{\textbf{(B)}\ -\frac{2\sqrt{3}}{3}}$ 是一个可能的首项。

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