1. **State the problem:** We have a loan of 9500 to be repaid with monthly payments of 325 at the beginning of each month. The interest rate is 9.50% compounded monthly. We need to find: a) The number of payments required to repay the debt, rounding up to the next whole payment. b) The size of the final payment, rounded to the nearest cent. 2. **Formula and important rules:** Since payments are at the beginning of each period, this is an annuity due. The present value $PV$ of an annuity due is given by: $$PV = P \times \frac{1 - (1 + i)^{-n}}{i} \times (1 + i)$$ where: - $P$ is the payment amount per period, - $i$ is the monthly interest rate (annual rate divided by 12), - $n$ is the number of payments. We know $PV = 9500$, $P = 325$, and $i = \frac{9.5}{100 \times 12} = 0.0079167$ approximately. 3. **Find the number of payments $n$:** Rearranging the formula: $$\frac{PV}{P(1+i)} = \frac{1 - (1+i)^{-n}}{i}$$ Multiply both sides by $i$: $$i \times \frac{PV}{P(1+i)} = 1 - (1+i)^{-n}$$ Then: $$(1+i)^{-n} = 1 - i \times \frac{PV}{P(1+i)}$$ Take natural logarithm: $$-n \ln(1+i) = \ln\left(1 - i \times \frac{PV}{P(1+i)}\right)$$ Solve for $n$: $$n = - \frac{\ln\left(1 - i \times \frac{PV}{P(1+i)}\right)}{\ln(1+i)}$$ 4. **Calculate intermediate values:** $$i = \frac{9.5}{100 \times 12} = 0.0079167$$ $$1+i = 1.0079167$$ $$\frac{PV}{P(1+i)} = \frac{9500}{325 \times 1.0079167} = \frac{9500}{327.5729} \approx 28.99$$ $$i \times \frac{PV}{P(1+i)} = 0.0079167 \times 28.99 \approx 0.2294$$ $$1 - 0.2294 = 0.7706$$ $$\ln(0.7706) = -0.2607$$ $$\ln(1+i) = \ln(1.0079167) = 0.007885$$ 5. **Calculate $n$:** $$n = - \frac{-0.2607}{0.007885} = 33.06$$ Rounding up, the number of payments required is **34**. 6. **Calculate the size of the final payment:** The total amount paid after 33 full payments is: $$A = P \times \frac{(1+i)^n - 1}{i} \times (1+i)^{-1}$$ But since payments are at the beginning, the balance after 33 payments is: $$B = PV \times (1+i)^{33} - P \times \frac{(1+i)^{33} - 1}{i}$$ Calculate $B$ to find the remaining balance before the 34th payment. Calculate: $$(1+i)^{33} = 1.0079167^{33} = e^{33 \times 0.007885} = e^{0.2602} = 1.2977$$ $$PV \times (1+i)^{33} = 9500 \times 1.2977 = 12328.15$$ $$\frac{(1+i)^{33} - 1}{i} = \frac{1.2977 - 1}{0.0079167} = \frac{0.2977}{0.0079167} = 37.60$$ $$P \times \frac{(1+i)^{33} - 1}{i} = 325 \times 37.60 = 12220$$ $$B = 12328.15 - 12220 = 108.15$$ So the final payment is approximately **108.15**. **Final answers:** - Number of payments required: **34** - Size of the final payment: **108.15**