Answer:
[tex]\textsf{A)} \quad -\dfrac{9}{5}[/tex]
Step-by-step explanation:
[tex]\boxed{\begin{minipage}{5.5 cm}\underline{Sum of an infinite geometric series}\\\\$S_{\infty}=\dfrac{a}{1-r}$\\\\where:\\\phantom{ww}$\bullet$ $a$ is the first term. \\ \phantom{ww}$\bullet$ $r$ is the common ratio.\\\end{minipage}}[/tex]
Given sum:
[tex]\displaystyle \sum^{\infty}_{n=1} -3\left(\dfrac{3}{8}\right)^n[/tex]
Therefore, the first term in the series is:
[tex]a_1=-3\left(\dfrac{3}{8}\right)^1=-\dfrac{9}{8}[/tex]
The common ratio, r, is:
[tex]r=\dfrac{3}{8}[/tex]
Substitute the first term, a, and common ratio, r, into the formula:
[tex]\implies S_{\infty}=\dfrac{-\frac{9}{8}}{1-\frac{3}{8}}[/tex]
[tex]\implies S_{\infty}=\dfrac{-\frac{9}{8}}{\frac{5}{8}}[/tex]
[tex]\implies S_{\infty}=-\dfrac{9}{5}[/tex]
Therefore, the sum of the given series is:
