1. **State the problem:** Determine the general term and the 10th term ($t_{10}$) of the sequence: 80, 40, 20, 10, ... 2. **Identify the sequence type:** Check if the sequence is arithmetic or geometric. - Arithmetic sequence has a constant difference: $d = t_{n} - t_{n-1}$. - Geometric sequence has a constant ratio: $r = \frac{t_{n}}{t_{n-1}}$. Calculate the difference and ratio: $$d = 40 - 80 = -40$$ $$r = \frac{40}{80} = \frac{1}{2}$$ Check next terms: $$20 - 40 = -20 \neq -40$$ (not constant difference) $$\frac{20}{40} = \frac{1}{2}$$ (constant ratio) So, the sequence is geometric with common ratio $r = \frac{1}{2}$. 3. **General term formula for geometric sequence:** $$t_n = t_1 \times r^{n-1}$$ where $t_1 = 80$ and $r = \frac{1}{2}$. 4. **Write the general term:** $$t_n = 80 \times \left(\frac{1}{2}\right)^{n-1}$$ 5. **Calculate $t_{10}$:** $$t_{10} = 80 \times \left(\frac{1}{2}\right)^{10-1} = 80 \times \left(\frac{1}{2}\right)^9 = 80 \times \frac{1}{512} = \frac{80}{512}$$ Simplify fraction: $$\frac{80}{512} = \frac{\cancel{80}^{16}}{\cancel{512}^{1024}} = \frac{5}{32}$$ 6. **Final answer:** - The sequence is geometric with common ratio $r = \frac{1}{2}$. - General term: $$t_n = 80 \times \left(\frac{1}{2}\right)^{n-1}$$ - $t_{10} = \frac{5}{32}$