The formula for the nth term of an arithmetic progression is given by:

\(a + (n-1) imes 3 = 94\)

For the sum of the first n terms, we use:

\(\frac{n}{2} [2a + (n-1) imes 3] = 1420\)

Substituting the first equation into the second:

\(\frac{n}{2} [a + 94] = 1420\)

Attempt elimination of a or n:

\(3n^2 - 191n + 2840 = 0\)

Solving this quadratic equation gives:

\(n = 40\)

Substitute back to find a:

\(a + 39 imes 3 = 94\)

\(a + 117 = 94\)

\(a = -23\)