(i) Sequence: ( 100, 100\times 0.8, 100\times 0.8^2, 100\times 0.8^3, \ldots ) [ a_n = 100 \times (0.8)^{n-1}. ]

(ii) Sum of first 15 terms of the G.P.: [ S_{15} = 100 \cdot \frac{1 - (0.8)^{15}}{1 - 0.8} = 100 \cdot \frac{1 - (0.8)^{15}}{0.2} = 500\bigl(1 - (0.8)^{15}\bigr). ] [ (0.8)^{15} \approx 0.035184,\quad S_{15} \approx 500(1 - 0.035184) = 500\times 0.964816 = 482.408. ]

(iii) Maximum (sum to infinity): [ S_\infty = \frac{100}{1-0.8} = \frac{100}{0.2} = 500. ]

Answers:
(i) ( a_n = 100(0.8)^{n-1} )
(ii) ( \approx 482.41 ) vehicles
(iii) 500 vehicles