Question

一个等比数列的第二项和第四项分别是 $2$ 和 $6$ 。下列哪一项可能是第一项？

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

$-\sqrt{3}$

$-\dfrac{2\sqrt{3}}{3}$

$-\dfrac{\sqrt{3}}{3}$

$\sqrt{3}$

$3$

若第一项为 $a$ ，公比为 $r$ ，则 $ar = 2$ 且 $ar^3 = 6$

If the first term is $a$ and the ratio is $r,$ then $ar = 2$ and $ar^3 = 6$

两式相除得到 $r^2 = 3$ ，所以 $r = \pm\sqrt{3}$

Dividing the two equations gives $r^2 = 3,$ so $r = \pm\sqrt{3}$

设第一项为 $a$ ，公比为 $r$ ，则 $ar = 2$ ，且 $ar^3 = 6$ ，因此 $r^2 = 3$ ， $r = \pm\sqrt{3}$ 。

第一项为 $a = \frac{2}{r} = \pm\frac{2}{\sqrt{3}} = \pm\frac{2\sqrt{3}}{3}\text{。}$ 选项中出现的是 $-\dfrac{2\sqrt{3}}{3}$ 。

因此，正确答案是 B。

Let the first term be $a$ and the common ratio $r.$ Then $ar = 2$ and $ar^3 = 6,$ so $r^2 = 3$ and $r = \pm\sqrt{3}.$

The first term is $a = \frac{2}{r} = \pm\frac{2}{\sqrt{3}} = \pm\frac{2\sqrt{3}}{3}.$ The choice $-\dfrac{2\sqrt{3}}{3}$ appears among the options.

Thus, the correct answer is B.