1. **State the problem:** Alisha invests 2000 into an account with varying effective monthly interest rates over three years. We need to find the amount after three years and the equivalent effective annual interest rate for the entire period. 2. **Given data:** - Initial principal $P = 2000$ - Interest rates: - First 6 months: $0.10\%$ per month - Next 24 months (2 years): $0.25\%$ per month - Last 6 months: $0.20\%$ per month 3. **Formula for compound interest with monthly compounding:** $$ A = P \times (1 + i_1)^{n_1} \times (1 + i_2)^{n_2} \times (1 + i_3)^{n_3} $$ where $i_k$ is the monthly interest rate for period $k$ and $n_k$ is the number of months in period $k$. 4. **Calculate each factor:** - First 6 months: $i_1 = 0.0010$, $n_1 = 6$ - Next 24 months: $i_2 = 0.0025$, $n_2 = 24$ - Last 6 months: $i_3 = 0.0020$, $n_3 = 6$ 5. **Calculate amount after 3 years:** $$ A = 2000 \times (1 + 0.0010)^6 \times (1 + 0.0025)^{24} \times (1 + 0.0020)^6 $$ 6. **Calculate each term:** $$ (1 + 0.0010)^6 = 1.006015 \quad (1 + 0.0025)^{24} = 1.061678 \quad (1 + 0.0020)^6 = 1.012072 $$ 7. **Multiply all terms:** $$ A = 2000 \times 1.006015 \times 1.061678 \times 1.012072 = 2000 \times 1.0819 = 2163.80 $$ 8. **Equivalent effective annual interest rate:** The total period is 3 years, so $$ (1 + r_{eff})^3 = \frac{A}{P} = \frac{2163.80}{2000} = 1.0819 $$ 9. **Solve for $r_{eff}$:** $$ r_{eff} = \sqrt[3]{1.0819} - 1 = 1.0264 - 1 = 0.0264 = 2.64\% $$ **Final answers:** - Amount after 3 years: $2163.80$ - Equivalent effective annual interest rate: $2.64\%$