1. **Problem statement:** Find the sum of the first $n$ terms and the sum to infinity of a geometric sequence where the first term $a_1=18$ and the fourth term $a_4=-\frac{2}{3}$. 2. **Recall the formula for the $n$th term of a geometric sequence:** $$a_n = a_1 r^{n-1}$$ where $r$ is the common ratio. 3. **Find the common ratio $r$ using the fourth term:** $$a_4 = a_1 r^{3} = -\frac{2}{3}$$ Substitute $a_1=18$: $$18 r^{3} = -\frac{2}{3}$$ Divide both sides by 18: $$\cancel{18} r^{3} = -\frac{2}{3} \div \cancel{18}$$ $$r^{3} = -\frac{2}{3} \times \frac{1}{18} = -\frac{2}{54} = -\frac{1}{27}$$ 4. **Solve for $r$:** $$r = \sqrt[3]{-\frac{1}{27}} = -\frac{1}{3}$$ 5. **Sum of the first $n$ terms formula:** $$S_n = a_1 \frac{1-r^n}{1-r}$$ Substitute $a_1=18$ and $r=-\frac{1}{3}$: $$S_n = 18 \frac{1-\left(-\frac{1}{3}\right)^n}{1 - \left(-\frac{1}{3}\right)}$$ 6. **Sum to infinity formula (valid if $|r|<1$):** $$S_\infty = \frac{a_1}{1-r}$$ Since $|r|=\frac{1}{3}<1$, sum to infinity exists. 7. **Calculate sum to infinity:** $$S_\infty = \frac{18}{1 - \left(-\frac{1}{3}\right)} = \frac{18}{1 + \frac{1}{3}} = \frac{18}{\frac{4}{3}} = 18 \times \frac{3}{4} = \frac{54}{4} = \frac{27}{2} = 13.5$$ **Final answers:** - Sum of first $n$ terms: $$S_n = 18 \frac{1-\left(-\frac{1}{3}\right)^n}{1 + \frac{1}{3}} = 18 \frac{1-\left(-\frac{1}{3}\right)^n}{\frac{4}{3}} = \frac{54}{4} \left(1-\left(-\frac{1}{3}\right)^n\right) = \frac{27}{2} \left(1-\left(-\frac{1}{3}\right)^n\right)$$ - Sum to infinity: $$S_\infty = \frac{27}{2} = 13.5$$