1. **State the problem:** Find the sum of the first 35 terms of the geometric series: $$8, -12, 18, \ldots$$ 2. **Identify the first term and common ratio:** The first term is $$a = 8$$. The common ratio $$r$$ is found by dividing the second term by the first term: $$r = \frac{-12}{8} = -\frac{3}{2}$$. 3. **Formula for the sum of the first $$n$$ terms of a geometric series:** $$S_n = a \frac{1 - r^n}{1 - r}$$ 4. **Apply the formula:** $$S_{35} = 8 \times \frac{1 - \left(-\frac{3}{2}\right)^{35}}{1 - \left(-\frac{3}{2}\right)}$$ 5. **Simplify the denominator:** $$1 - \left(-\frac{3}{2}\right) = 1 + \frac{3}{2} = \frac{5}{2}$$ 6. **Calculate the numerator:** $$1 - \left(-\frac{3}{2}\right)^{35}$$ Since $$\left(-\frac{3}{2}\right)^{35} = - \left(\frac{3}{2}\right)^{35}$$ (because 35 is odd), $$1 - \left(-\frac{3}{2}\right)^{35} = 1 - (-X) = 1 + X$$ where $$X = \left(\frac{3}{2}\right)^{35}$$. 7. **Rewrite the sum:** $$S_{35} = 8 \times \frac{1 + \left(\frac{3}{2}\right)^{35}}{\frac{5}{2}} = 8 \times \frac{2}{5} \times \left(1 + \left(\frac{3}{2}\right)^{35}\right) = \frac{16}{5} \left(1 + \left(\frac{3}{2}\right)^{35}\right)$$ 8. **Calculate $$\left(\frac{3}{2}\right)^{35}$$ approximately:** $$\left(\frac{3}{2}\right)^{35} = 1.5^{35} \approx 1667986.995$$ (using a calculator). 9. **Calculate the sum:** $$S_{35} \approx \frac{16}{5} \times (1 + 1667986.995) = \frac{16}{5} \times 1667987.995 = 3.2 \times 1667987.995 \approx 5337561$$ **Final answer:** The sum of the first 35 terms is approximately $$5337561$$ (a whole number).