Answer:
[tex]S_{\infty}=-\dfrac{625}{396}[/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 geometric series:
[tex]-\dfrac{125}{36}+\dfrac{25}{6}-5+6-...[/tex]
To find the common ratio, divide a term by the previous term:
[tex]\implies r=\dfrac{a_4}{a_3}=-\dfrac{6}{5}[/tex]
Substitute the found common ratio and given first term into the sum formula:
[tex]\implies S_{\infty}=\dfrac{-\frac{125}{36}}{1-\left(-\frac{6}{5}\right)}[/tex]
[tex]\implies S_{\infty}=\dfrac{-\frac{125}{36}}{1+\frac{6}{5}}[/tex]
[tex]\implies S_{\infty}=\dfrac{-\frac{125}{36}}{\frac{11}{5}}[/tex]
[tex]\implies S_{\infty}=-\dfrac{125}{36} \times\dfrac{5}{11}[/tex]
[tex]\implies S_{\infty}=-\dfrac{625}{396}[/tex]
