1. **State the problem:** We have an arithmetic progression (AP) where the first term $a_1 = 300$ metres and the common difference $d = 50$ metres. The total number of rounds is 10. 2. **Find the 4th, 5th, and 6th terms of the AP:** The formula for the $n$th term of an AP is: $$a_n = a_1 + (n-1)d$$ Calculate each term: - $a_4 = 300 + (4-1) \times 50 = 300 + 150 = 450$ - $a_5 = 300 + (5-1) \times 50 = 300 + 200 = 500$ - $a_6 = 300 + (6-1) \times 50 = 300 + 250 = 550$ 3. **Determine the distance of the 8th round:** Using the same formula: $$a_8 = 300 + (8-1) \times 50 = 300 + 350 = 650$$ 4. **Find the total distance run after completing all 10 rounds:** The sum of the first $n$ terms of an AP is: $$S_n = \frac{n}{2} (2a_1 + (n-1)d)$$ Calculate $S_{10}$: $$S_{10} = \frac{10}{2} (2 \times 300 + (10-1) \times 50) = 5 (600 + 450) = 5 \times 1050 = 5250$$ 5. **Find the total distance run if only the first 6 rounds are completed:** Calculate $S_6$: $$S_6 = \frac{6}{2} (2 \times 300 + (6-1) \times 50) = 3 (600 + 250) = 3 \times 850 = 2550$$ **Final answers:** - Fourth term: $450$ metres - Fifth term: $500$ metres - Sixth term: $550$ metres - Distance of 8th round: $650$ metres - Total distance after 10 rounds: $5250$ metres - Total distance after 6 rounds: $2550$ metres