Chapter: Arithmetic Progression· Mathematics Part 1 (Algebra)
Answer: 20 weeks
\(a = 50\), \(d = 10\), \(S_n = 2900\)
\(2900 = \dfrac{n}{2}[100 + (n - 1) \times 10] = \dfrac{n}{2}(90 + 10n) = 5n(9 + n)\)
\(n^2 + 9n = 580 \Rightarrow n^2 + 9n - 580 = 0\)
\((n + 29)(n - 20) = 0 \Rightarrow n = 20\) (weeks cannot be negative).
Check: \(t_{20} = 50 + 190 = 240\); \(S_{20} = 10(50 + 240) = 2900\) ✓.