Q1. [4]
A school is organizing a charity run to raise funds for a local hospital. The run is planned as a series of rounds around a track, with each round being 300 metres. To make the event more challenging and engaging, the organizers decide to increase the distance of each subsequent round by 50 metres. For example, the second round will be 350 metres, the third round will be 400 metres and so on. The total number of rounds planned is 10.
Based on the information given above, answer the following questions:
- (i) Write the fourth, fifth and sixth term of the Arithmetic Progression so formed. [1]
- (ii) Determine the distance of the 8th round. [1]
- (iii) Find the total distance run after completing all 10 rounds. [2]
Previously asked in CBSE board exam
2025 30/1/1 Q36
Generated by claude-sonnet-4-6 · 2026-06-15 10:30 · grounding stimulus
Model Answer
The AP formed is: 300, 350, 400, 450, 500, 550, …
Here, first term $a = 300$ and common difference $d = 50$.
(i) 4th term = 450 m, 5th term = 500 m, 6th term = 550 m
(ii) Distance of 8th round:
$a_8 = a + 7d = 300 + 7 \times 50 = 300 + 350 = \mathbf{650}$ metres
(iii) Total distance after 10 rounds:
$$S_{10} = \frac{n}{2}[2a + (n-1)d] = \frac{10}{2}[2 \times 300 + 9 \times 50]$$
$$= 5 \times [600 + 450] = 5 \times 1050 = \mathbf{5250 \text{ metres}}$$
Source: Arithmetic Progressions, AP Sum Formula
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Explanation
- Examiner focus: Correct identification of $a$ and $d$ is key; errors here lose marks throughout.
- (i) Simply list terms — no formula needed.
- (ii) Use $a_n = a + (n-1)d$; substitution must be shown for full credit.
- (iii) Must write the $S_n$ formula, substitute values, and show working step-by-step — both steps carry marks. Final answer in metres.
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