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

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

For the first 200 terms:

\(S_{200} = \frac{200}{2} (2a + 199d) = 100(2a + 199d)\)

For the first 100 terms:

\(S_{100} = \frac{100}{2} (2a + 99d) = 50(2a + 99d)\)

Given that \(S_{200} = 4S_{100}\):

\(100(2a + 199d) = 4 \times 50(2a + 99d)\)

\(100(2a + 199d) = 200(2a + 99d)\)

\(2a + 199d = 2(2a + 99d)\)

\(2a + 199d = 4a + 198d\)

\(199d - 198d = 4a - 2a\)

\(d = 2a\)

For the 100th term:

The 100th term \(T_{100}\) is given by:

\(T_{100} = a + 99d\)

Substituting \(d = 2a\):

\(T_{100} = a + 99(2a)\)

\(T_{100} = a + 198a\)

\(T_{100} = 199a\)