1. **Problem statement:** A competitor runs a 20 km race. - First 4 km at 6 minutes per km. - After 4 km, each subsequent km takes 5% longer than the previous km. 2. **Part (a): Show total time for first 6 km is 36 minutes 55 seconds.** - Time for first 4 km: $4 \times 6 = 24$ minutes. - Time for 5th km: $6 \times 1.05 = 6.3$ minutes. - Time for 6th km: $6.3 \times 1.05 = 6 \times 1.05^2 = 6 \times 1.1025 = 6.615$ minutes. - Total time for first 6 km: $$24 + 6.3 + 6.615 = 36.915 \text{ minutes}$$ - Convert $0.915$ minutes to seconds: $$0.915 \times 60 = 54.9 \approx 55 \text{ seconds}$$ - So total time is approximately $36$ minutes $55$ seconds. 3. **Part (b): Show time for kth km ($5 \leq k \leq 20$) is $6 \times 1.05^{k-4}$.** - Time for 5th km: $6 \times 1.05^{5-4} = 6 \times 1.05^1 = 6.3$ minutes. - Time for 6th km: $6 \times 1.05^{6-4} = 6 \times 1.05^2 = 6.615$ minutes. - Time for 7th km: $6 \times 1.05^{7-4} = 6 \times 1.05^3 = 6 \times 1.157625 = 6.94575$ minutes. - This matches the pattern of increasing time by 5% each km after the 4th. Hence, the formula for the time to run the kth km for $k \geq 5$ is: $$t_k = 6 \times 1.05^{k-4}$$ **Final answers:** - (a) $36$ minutes $55$ seconds - (b) $t_k = 6 \times 1.05^{k-4}$ for $5 \leq k \leq 20$