Answer:
General term of the sequence will be [tex]a_{n}=-8\left ( \dfrac{-2}{3}\right )^{n}[/tex].
Step-by-step explanation:
The given sequence is [tex]-8,\dfrac{16}{3}, \dfrac{-32}{9}, \dfrac{64}{27},\dfrac{-128}{81},.....[/tex].
In the given sequence, [tex]-8[/tex] is the first term. The sequence is a geometric progression. So, the common ratio of the sequence will be,
[tex]r=\dfrac{16}{3}\times\dfrac{1}{-8}\\r=\dfrac{-2}{3}[/tex]
Now, the nth term of the sequence will be,
[tex]a_{n}=ar^{n}\\a_{n}=-8\times\left ( \dfrac{-2}{3}\right )^{n} \\a_{n}=-8\left ( \dfrac{-2}{3}\right )^{n}[/tex]
where n=1, 2, 3,....
For more details, refer the link:
https://brainly.com/question/4853032?referrer=searchResults
