1. **Stating the problem:** We want to find the amount in the account immediately before the 6th year's investment, with R5601 invested every 2 months, at an annual interest rate of 10.131%. The first investment is at time 0. 2. **Identify parameters:** - Investment amount per period, $P = 5601$ - $n = 6$ years - Compounding/investment frequency: every 2 months, so 6 times per year - Annual interest rate: $i = 10.131\% = 0.10131$ 3. **Calculate the periodic interest rate:** Since 6 periods per year, $$ i_p = \frac{0.10131}{6} = 0.016885\text{ (approximately)} $$ 4. **Calculate total number of investments/periods:** $$ N = 6 \times 6 = 36 $$ 5. **Use the formula for the future value of an annuity with payments at the beginning of each period (annuity due):** $$ FV = P \times \frac{(1 + i_p)^N - 1}{i_p} \times (1 + i_p) $$ 6. **Substitute values:** $$ FV = 5601 \times \frac{(1 + 0.016885)^{36} - 1}{0.016885} \times (1 + 0.016885) $$ 7. **Calculate $(1 + 0.016885)^{36}$:** $$ (1.016885)^{36} \approx 1.8119 $$ 8. **Calculate numerator:** $$ 1.8119 - 1 = 0.8119 $$ 9. **Calculate fraction:** $$ \frac{0.8119}{0.016885} \approx 48.09 $$ 10. **Calculate full expression:** $$ FV = 5601 \times 48.09 \times 1.016885 \approx 5601 \times 48.9 = 273850.61 $$ 11. **Rounding to 2 decimal places:** $$ FV \approx 273850.61 $$ **Final answer:** The amount in the account immediately before the investment on the 6th year day is approximately **273850.61**.