Given:
[tex]-12-4-\frac{4}{3}-\ldots\text{..-}\frac{4}{243}[/tex]
Find the general representation of the given series,
[tex]\begin{gathered} a_n=ar^{n-1} \\ a=\text{ first term =-12} \\ r=\text{ common ratio = 1/3} \\ a_n=-12(\frac{1}{3})^{n-1} \end{gathered}[/tex]
As, the nth term of the series is -4/243.
[tex]\begin{gathered} a_n=-12(\frac{1}{3})^{n-1} \\ -\frac{4}{243}_{}=-12(\frac{1}{3})^{n-1} \\ \frac{1}{729}=(\frac{1}{3})^{n-1} \\ \text{Apply exponent rule} \\ -(n-1)=-6 \\ n=7 \end{gathered}[/tex]
The sum of the first 7 terms of the given series is,
[tex]\begin{gathered} S_n=\frac{a(1-r^n)}{1-r},|r|<1 \\ S_7=\frac{-12(1-(\frac{1}{3})^7}{1-\frac{1}{3}} \\ =\frac{-\frac{8744}{729}}{\frac{2}{3}} \\ =-\frac{4372}{243} \\ =-17.99 \\ \approx-18 \end{gathered}[/tex]
Answer: the sum of the series is -18.
