(i) The third term is given by \(a + 2d = 90\) and the fifth term by \(a + 4d = 80\). Solving these equations simultaneously:

Subtract the first from the second: \((a + 4d) - (a + 2d) = 80 - 90\)

\(2d = -10\)

\(d = -5\)

Substitute \(d = -5\) into \(a + 2d = 90\):

\(a + 2(-5) = 90\)

\(a - 10 = 90\)

\(a = 100\)

(ii) The sum of the first \(m\) terms is equal to the sum of the first \((m + 1)\) terms:

\(S_m = S_{m+1}\)

\(\frac{m}{2}(2a + (m-1)d) = \frac{m+1}{2}(2a + md)\)

Since \(S_m = S_{m+1}\), the difference is zero:

\(a + md = 0\)

\(100 + m(-5) = 0\)

\(100 - 5m = 0\)

\(5m = 100\)

\(m = 20\)

(iii) The sum of the first \(n\) terms is zero:

\(S_n = 0\)

\(\frac{n}{2}(2a + (n-1)d) = 0\)

\(n(200 + (n-1)(-5)) = 0\)

\(n(200 - 5n + 5) = 0\)

\(n(205 - 5n) = 0\)

\(205 - 5n = 0\)

\(5n = 205\)

\(n = 41\)