In order to organise, Annual Sports Day, a school prepared an eight lane running track with an integrated football field inside the track area as shown below. The length of innermost lane of the track is 400 m and each subsequent lane is 7.6 m longer than the preceding lane.
Based on given information, answer the following questions, using concept of Arithmetic Progression.
Generated by claude-sonnet-4-6 · 2026-06-15 10:33 · grounding stimulus
Model Answer
Given: a = 400 m, d = 7.6 m (AP of lane lengths)
(i) Length of 6th lane:
$a_6 = a + 5d = 400 + 5(7.6) = 400 + 38 = \mathbf{438\ m}$
(ii) How much longer is 8th lane than 4th lane:
$a_8 = 400 + 7(7.6) = 400 + 53.2 = 453.2\ m$
$a_4 = 400 + 3(7.6) = 400 + 22.8 = 422.8\ m$
Difference $= 453.2 - 422.8 = \mathbf{30.4\ m}$
(iii) Total distance covered in first six lanes:
$S_6 = \frac{6}{2}[2a + 5d] = 3[2(400) + 5(7.6)] = 3[800 + 38] = 3 \times 838 = \mathbf{2514\ m}$
Source: Arithmetic Progressions, Application-based case study
---
Explanation
- Lane lengths form an AP with first term a = 400, common difference d = 7.6.
- For (ii), the difference between any two lanes = (difference in their positions) × d = 4 × 7.6 = 30.4 m — you can use this shortcut.
- For (iii), use sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$ with n = 6. Examiners award 1 mark for correct formula and 1 mark for correct calculation.