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

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

where \(a\) is the first term and \(d\) is the common difference.

Given: \(a = 7\), \(a_n = 84\), and \(a_{3n} = 245\).

For the nth term:

\(7 + (n-1)d = 84\)

\((n-1)d = 77\)

For the (3n)th term:

\(7 + (3n-1)d = 245\)

\((3n-1)d = 238\)

Now, divide the two equations:

\(\frac{n-1}{3n-1} = \frac{77}{238}\)

Simplify \(\frac{77}{238}\) to \(\frac{11}{34}\):

\(\frac{n-1}{3n-1} = \frac{11}{34}\)

Cross-multiply:

\(34(n-1) = 11(3n-1)\)

\(34n - 34 = 33n - 11\)

\(n = 23\)