(i) The distance cycled on the nth day is given by the formula for the nth term of an arithmetic sequence:

\(a_n = a + (n-1) imes d\)

where \(a = 200\) km (the distance on the first day) and \(d = -5\) km (the reduction each day). For May 15th, \(n = 15\):

\(a_{15} = 200 + (15-1) imes (-5)\)

\(a_{15} = 200 - 70 = 130\) km

(ii) The total distance cycled is the sum of an arithmetic series:

\(S_n = \frac{n}{2} [2a + (n-1) imes d]\)

We need \(S_n = 3050\) km, \(a = 200\), and \(d = -5\):

\(\frac{n}{2} [400 + (n-1) imes (-5)] = 3050\)

\(\frac{n}{2} [400 - 5n + 5] = 3050\)

\(n(405 - 5n) = 6100\)

\(5n^2 - 405n + 6100 = 0\)

Solving this quadratic equation gives \(n = 20\).

Therefore, he finishes on May 20th.