1. **Problem Statement:** A pair of parents start investing $2000 in year 1 and increase the deposit by 10% each year. (a) Find the amount invested in the 19th year. (b) Find the total amount invested over 19 years. 2. **Formula and Explanation:** This is a geometric sequence where the first term $a_1 = 2000$ and the common ratio $r = 1 + 0.10 = 1.10$. - The amount invested in year $n$ is given by: $$a_n = a_1 \times r^{n-1}$$ - The total amount invested over $n$ years is the sum of the geometric series: $$S_n = a_1 \times \frac{r^n - 1}{r - 1}$$ 3. **Calculate the amount invested in the 19th year:** $$a_{19} = 2000 \times 1.10^{18}$$ Calculate $1.10^{18}$: $$1.10^{18} \approx 5.5599$$ So, $$a_{19} = 2000 \times 5.5599 = 11119.8$$ 4. **Calculate the total amount invested over 19 years:** $$S_{19} = 2000 \times \frac{1.10^{19} - 1}{1.10 - 1}$$ Calculate $1.10^{19}$: $$1.10^{19} \approx 6.1159$$ So, $$S_{19} = 2000 \times \frac{6.1159 - 1}{0.10} = 2000 \times \frac{5.1159}{0.10} = 2000 \times 51.159 = 102318$$ 5. **Final answers:** - Amount invested in the 19th year: **11119.8** - Total amount invested over 19 years: **102318**