Answer:
Yes, the sequence is geometric.
[tex]\textsf{Common ratio}=\dfrac{2\sqrt{3}}{3}[/tex]
Step-by-step explanation:
Given sequence:
[tex]2, \; \dfrac{4}{\sqrt{3}},\; \dfrac{8}{3},\;\dfrac{16}{3\sqrt{3}},\;...[/tex]
A geometric sequence has a common ratio.
Therefore, to check if the given sequence has a common ratio, divide each term by the previous term:
[tex]\boxed{\begin{aligned}\dfrac{16}{3\sqrt{3}} \div \dfrac{8}{3}&=\dfrac{16}{3\sqrt{3}} \times \dfrac{3}{8}\\\\&=\dfrac{48}{24\sqrt{3}}\\\\&=\dfrac{2}{\sqrt{3}}\\\\&=\dfrac{2\sqrt{3}}{3}\end{aligned}}[/tex]
[tex]\boxed{\begin{aligned}\dfrac{8}{3} \div \dfrac{4}{\sqrt{3}}&=\dfrac{8}{3} \times \dfrac{\sqrt{3}}{4}\\\\&=\dfrac{8\sqrt{3}}{12}\\\\&=\dfrac{2\sqrt{3}}{3}\end{aligned}}[/tex]
[tex]\boxed{\begin{aligned} \dfrac{4}{\sqrt{3}} \div 2&= \dfrac{4}{\sqrt{3}} \times \dfrac{1}{2}\\\\&= \dfrac{4}{2\sqrt{3}} \\\\&=\dfrac{2}{\sqrt{3}}\\\\&=\dfrac{2\sqrt{3}}{3}\end{aligned}}[/tex]
As there is a common ratio, the sequence is geometric.
The common ratio is:
- [tex]\dfrac{2\sqrt{3}}{3}[/tex]
