Given that the sum to infinity \(s,=a+a r+a r^{2}\) \(=\frac{a}{1-r}\). to what sum does the infinity series \(1-\frac{2}{3}+\frac{4}{9}-\frac{8}{27}+\ldots\) converge?
A. \(-\frac{3}{5} \) B. \(\frac{5}{3}\) C. \(-\frac{5}{3}\) D. \(\frac{3}{5}\)
Correct Answer: D
Explanation
given \begin{aligned} S_{x} &=a+c r+a r^{2}+\ldots=\frac{a}{1-r} \\ & 1-\frac{2}{3}+\frac{4}{9}-\frac{8}{27}+\ldots \\ \text { first term } a &=1 \\ \text { common ratio }(r) &=\frac{\text { second term }}{\text { first term }} \\ &=\frac{-2}{\frac{3}{1}}=\frac{-2}{3} \\ S_{,} &=\frac{a}{1-r}=\frac{1}{1-\left(\frac{-2}{3}\right)}=\frac{1}{1+\frac{2}{3}} \\ &=\frac{1}{5}=\frac{3}{5} \end{aligned}