(a) The sum of the first \(n\) terms \(S_n\) of an arithmetic progression is given by:

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

Substituting the given values, \(a = 4\) and \(S_n = 5863\):

\(\frac{n}{2} [8 + (n-1)d] = 5863\)

Multiply both sides by 2:

\(n[8 + (n-1)d] = 11726\)

Divide by \(n\):

\(8 + (n-1)d = \frac{11726}{n}\)

Rearrange to find \((n-1)d\):

\((n-1)d = \frac{11726}{n} - 8\)

(b) The \(n\)th term \(u_n\) is given by:

\(u_n = a + (n-1)d\)

Given \(u_n = 139\):

\(4 + (n-1)d = 139\)

Substitute \((n-1)d = \frac{11726}{n} - 8\):

\(\frac{11726}{n} - 8 = 135\)

\(\frac{11726}{n} = 143\)

\(n = \frac{11726}{143} = 82\)

Substitute \(n = 82\) back into \((n-1)d = \frac{11726}{n} - 8\):

\(81d = \frac{11726}{82} - 8\)

\(81d = 143 - 8\)

\(81d = 135\)

\(d = \frac{135}{81} = \frac{5}{3}\)