1. **Problem Statement:** Kanda receives payments starting at 90,000 now, increasing by 7,000 each year until the last payment of 160,000. The annual effective interest rate is 9%. Find the present value of these payments. 2. **Identify the payment pattern:** Payments are: 90,000 at time 0, 97,000 at time 1, 104,000 at time 2, 111,000 at time 3, ..., up to 160,000. 3. **Number of payments:** The payments increase by 7,000 each year. To find the number of payments $n$, solve: $$90,000 + 7,000 \times n = 160,000$$ $$7,000 \times n = 70,000$$ $$n = 10$$ Since the first payment is at time 0, payments occur at times $0,1,2,\ldots,10$, total 11 payments. 4. **Present value formula:** The present value (PV) of a payment $P_t$ at time $t$ with interest rate $i$ is: $$PV = \sum_{t=0}^{10} \frac{P_t}{(1+i)^t}$$ where $P_t = 90,000 + 7,000t$ and $i=0.09$. 5. **Calculate PV:** $$PV = \sum_{t=0}^{10} \frac{90,000 + 7,000t}{(1.09)^t} = \sum_{t=0}^{10} \frac{90,000}{(1.09)^t} + \sum_{t=0}^{10} \frac{7,000t}{(1.09)^t}$$ 6. **Calculate each sum separately:** - First sum is a geometric series: $$S_1 = 90,000 \sum_{t=0}^{10} (1.09)^{-t}$$ - Second sum is an arithmetico-geometric series: $$S_2 = 7,000 \sum_{t=0}^{10} t (1.09)^{-t}$$ 7. **Sum of geometric series:** $$\sum_{t=0}^{n} r^t = \frac{1-r^{n+1}}{1-r}$$ Here, $r = \frac{1}{1.09} \approx 0.917431$ and $n=10$. Calculate: $$\sum_{t=0}^{10} (1.09)^{-t} = \frac{1 - (1.09)^{-11}}{1 - (1.09)^{-1}} = \frac{1 - (1.09)^{-11}}{1 - 0.917431}$$ Calculate $(1.09)^{-11} = \frac{1}{1.09^{11}} \approx 0.42241$. So: $$\sum_{t=0}^{10} (1.09)^{-t} = \frac{1 - 0.42241}{1 - 0.917431} = \frac{0.57759}{0.082569} \approx 6.997$$ 8. **Sum of arithmetico-geometric series:** Formula: $$\sum_{t=0}^n t r^t = r \frac{1 - (n+1) r^n + n r^{n+1}}{(1-r)^2}$$ Using $r=0.917431$, $n=10$: Calculate: $$(n+1) r^n = 11 \times 0.917431^{10}$$ Calculate $r^{10} = (1.09)^{-10} = \frac{1}{1.09^{10}} \approx 0.46017$. So: $$(n+1) r^n = 11 \times 0.46017 = 5.0619$$ Calculate: $$n r^{n+1} = 10 \times 0.917431^{11} = 10 \times 0.42241 = 4.2241$$ Now: $$\sum_{t=0}^{10} t r^t = 0.917431 \times \frac{1 - 5.0619 + 4.2241}{(1 - 0.917431)^2} = 0.917431 \times \frac{0.1622}{0.082569^2}$$ Calculate denominator squared: $$0.082569^2 = 0.00682$$ So: $$\sum_{t=0}^{10} t r^t = 0.917431 \times \frac{0.1622}{0.00682} = 0.917431 \times 23.79 = 21.82$$ 9. **Calculate total PV:** $$PV = 90,000 \times 6.997 + 7,000 \times 21.82 = 629,730 + 152,740 = 782,470$$ 10. **Final answer:** The present value of the payments is approximately **782,470**.