1. **State the problem:** Calculate the monthly installment payments for a 30-year adjustable rate mortgage with a loan amount of 60000, initial interest rate of 10% for year 1, and an increased interest rate of 12% for year 2. Payments are monthly (12 per year), and the interest rate resets yearly. 2. **Formula for monthly mortgage payment:** $$\text{Payment} = P \times \frac{r(1+r)^n}{(1+r)^n - 1}$$ where: - $P$ = loan amount - $r$ = monthly interest rate (annual rate divided by 12) - $n$ = total number of payments (years \times payments per year) 3. **Calculate year 1 payment:** - $P = 60000$ - Annual interest rate = 10% = 0.10 - Monthly interest rate $r = \frac{0.10}{12} = 0.0083333$ - Total payments $n = 30 \times 12 = 360$ Calculate numerator: $$r(1+r)^n = 0.0083333 \times (1 + 0.0083333)^{360}$$ Calculate denominator: $$(1+r)^n - 1 = (1 + 0.0083333)^{360} - 1$$ Using approximation: $$(1 + 0.0083333)^{360} \approx (1.0083333)^{360} \approx e^{360 \times \ln(1.0083333)}$$ Calculate $\ln(1.0083333) \approx 0.008298$ So, $$e^{360 \times 0.008298} = e^{2.987} \approx 19.83$$ Now numerator: $$0.0083333 \times 19.83 = 0.16525$$ Denominator: $$19.83 - 1 = 18.83$$ Monthly payment year 1: $$60000 \times \frac{0.16525}{18.83} = 60000 \times 0.00877 = 526.2$$ 4. **Calculate year 2 payment:** - New annual interest rate = 12% = 0.12 - Monthly interest rate $r = \frac{0.12}{12} = 0.01$ - Remaining term after year 1 = 29 years = 348 payments Calculate numerator: $$0.01 \times (1 + 0.01)^{348}$$ Calculate denominator: $$(1 + 0.01)^{348} - 1$$ Approximate: $$\ln(1.01) \approx 0.00995$$ $$e^{348 \times 0.00995} = e^{3.46} \approx 31.8$$ Numerator: $$0.01 \times 31.8 = 0.318$$ Denominator: $$31.8 - 1 = 30.8$$ Monthly payment year 2: $$60000 \times \frac{0.318}{30.8} = 60000 \times 0.01032 = 619.2$$ **Final answers:** - Year 1 monthly installment = $526.20$ - Year 2 monthly installment = $619.20$