1. **State the problem:** We need to find the first five terms and the tenth term of a geometric sequence where the first term $a=3$ and the common ratio $r=\frac{-1}{2}$. 2. **Recall the formula for the $n^{th}$ term of a geometric sequence:** $$a_n = a \times r^{n-1}$$ where $a$ is the first term and $r$ is the common ratio. 3. **Calculate the first five terms:** - $a_1 = 3 \times \left(\frac{-1}{2}\right)^{0} = 3 \times 1 = 3$ - $a_2 = 3 \times \left(\frac{-1}{2}\right)^{1} = 3 \times \left(-\frac{1}{2}\right) = -\frac{3}{2} = -1.5$ - $a_3 = 3 \times \left(\frac{-1}{2}\right)^{2} = 3 \times \frac{1}{4} = \frac{3}{4} = 0.75$ - $a_4 = 3 \times \left(\frac{-1}{2}\right)^{3} = 3 \times \left(-\frac{1}{8}\right) = -\frac{3}{8} = -0.375$ - $a_5 = 3 \times \left(\frac{-1}{2}\right)^{4} = 3 \times \frac{1}{16} = \frac{3}{16} = 0.1875$ 4. **Calculate the tenth term:** $$a_{10} = 3 \times \left(\frac{-1}{2}\right)^{9}$$ Calculate the exponent: $$\left(\frac{-1}{2}\right)^{9} = \frac{-1^9}{2^9} = \frac{-1}{512}$$ Thus $$a_{10} = 3 \times \left(-\frac{1}{512}\right) = -\frac{3}{512} \approx -0.005859375$$ **Final answer:** - First five terms: $3$, $-1.5$, $0.75$, $-0.375$, $0.1875$ - Tenth term: $-\frac{3}{512}$ or approximately $-0.005859375$