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 20 terms:

\(\frac{20}{2} (2a + 19d) = 405\)

\(10(2a + 19d) = 405\)

For the first 40 terms:

\(\frac{40}{2} (2a + 39d) = 1410\)

\(20(2a + 39d) = 1410\)

We have the equations:

1. \(10(2a + 19d) = 405\)

2. \(20(2a + 39d) = 1410\)

Solving these simultaneously:

From equation 1: \(2a + 19d = 40.5\)

From equation 2: \(2a + 39d = 70.5\)

Subtract equation 1 from equation 2:

\(20d = 30\)

\(d = 1.5\)

Substitute \(d = 1.5\) into equation 1:

\(2a + 19(1.5) = 40.5\)

\(2a + 28.5 = 40.5\)

\(2a = 12\)

\(a = 6\)

Now, find the 60th term \(T_{60}\):

\(T_{60} = a + 59d\)

\(T_{60} = 6 + 59(1.5)\)

\(T_{60} = 6 + 88.5\)

\(T_{60} = 94.5\)