1. **Problem Statement:** Rorisang took a loan of 2500000 at 7.8% annual interest compounded monthly for 20 years. We need to find various loan payment details. 2. **Formula for monthly payment (PMT):** $$PMT = P \times \frac{r(1+r)^n}{(1+r)^n - 1}$$ where $P=2500000$, $r=\frac{0.078}{12}$ (monthly interest rate), $n=20 \times 12=240$ (total payments). 3. **Calculate monthly payment:** $r=0.0065$, $n=240$ $$PMT = 2500000 \times \frac{0.0065(1+0.0065)^{240}}{(1+0.0065)^{240} - 1}$$ Calculate $(1+0.0065)^{240} \approx 4.3496$ $$PMT = 2500000 \times \frac{0.0065 \times 4.3496}{4.3496 - 1} = 2500000 \times \frac{0.02827}{3.3496} = 2500000 \times 0.00844 = 21100$$ 4. **Interest in 135th payment:** Interest part = Outstanding principal before 135th payment $\times r$ Outstanding principal after 134 payments: $$B_k = P(1+r)^n - PMT \times \frac{(1+r)^n - (1+r)^k}{r}$$ For $k=134$: $$B_{134} = 2500000 \times 4.3496 - 21100 \times \frac{4.3496 - (1.0065)^{134}}{0.0065}$$ Calculate $(1.0065)^{134} \approx 2.456$ $$B_{134} = 10874000 - 21100 \times \frac{4.3496 - 2.456}{0.0065} = 10874000 - 21100 \times 292.32 = 10874000 - 6165992 = 4708008$$ Interest in 135th payment: $$I_{135} = 4708008 \times 0.0065 = 30602$$ 5. **Principal outstanding after 92nd payment:** For $k=92$: Calculate $(1.0065)^{92} \approx 1.819$ $$B_{92} = 2500000 \times 4.3496 - 21100 \times \frac{4.3496 - 1.819}{0.0065} = 10874000 - 21100 \times 388.63 = 10874000 - 8199643 = 2674357$$ 6. **Finance charge:** Total paid = $PMT \times n = 21100 \times 240 = 5064000$ Finance charge = Total paid - Principal = $5064000 - 2500000 = 2564000$ 7. **100th payment:** All payments are equal, so 100th payment = monthly payment = 21100 **Final answers:** - Monthly payment = 21100 - Interest in 135th payment = 30602 - Principal outstanding after 92nd payment = 2674357 - Finance charge = 2564000 - 100th payment = 21100