The minimum age of children eligible to participate in a painting competition is 8 years. It is observed that the age of the youngest boy was 8 years and the ages of the participants, when seated in order of age, have a common difference of 4 months. If the sum of the ages of all the participants is 168 years, find the age of the eldest participant in the painting competition.
Generated by claude-sonnet-4-6 · 2026-06-15 10:30 · grounding rag
Model Answer
Given: First term $a = 8$ years, common difference $d = 4$ months $= \dfrac{1}{3}$ year, Sum $S_n = 168$ years.
Using the formula:
$$S_n = \frac{n}{2}[2a + (n-1)d]$$
$$168 = \frac{n}{2}\left[2 \times 8 + (n-1) \times \frac{1}{3}\right]$$
$$168 = \frac{n}{2}\left[16 + \frac{n-1}{3}\right]$$
$$336 = n\left[\frac{48 + n - 1}{3}\right]$$
$$1008 = n(47 + n)$$
$$n^2 + 47n - 1008 = 0$$
$$n^2 + 63n - 16n - 1008 = 0$$
$$(n + 63)(n - 16) = 0$$
Since $n$ cannot be negative, $n = 16$.
Age of the eldest participant:
$$a_{16} = a + (n-1)d = 8 + 15 \times \frac{1}{3} = 8 + 5 = 13 \text{ years}$$
The age of the eldest participant is 13 years.
Source: Arithmetic Progressions, Section 5.4
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Explanation
- Convert $d = 4$ months to years ($\frac{1}{3}$ year) before substituting — a common error point.
- Substitute into $S_n = \frac{n}{2}[2a + (n-1)d]$, simplify to get a quadratic, and reject the negative root.
- Finally use $a_n = a + (n-1)d$ to find the last term. Examiners award marks at each step: identifying AP values (1 mark), setting up the sum equation (1 mark), solving the quadratic (2 marks), and finding the eldest age (1 mark).