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

A. \(-\sqrt{3}\)

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

C. \(-\frac{\sqrt{3}}{3}\)

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

 Jun 30, 2020
 #1
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Let the first term be $a$ and the common difference 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.

 Aug 28, 2020

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