Let the first term of the arithmetic progression be \(a = 60\) and the common difference be \(d\). The total number of payments is \(n = 48\), and the total amount repaid is \(S_n = 3726\).

The formula for the sum of an arithmetic progression is:

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

Substitute the given values:

\(3726 = \frac{48}{2} (2 \times 60 + 47d)\)

\(3726 = 24 (120 + 47d)\)

\(3726 = 2880 + 1128d\)

\(846 = 1128d\)

\(d = \frac{846}{1128} = 0.75\)

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

\(a_3 = a + 2d\)

\(a_3 = 60 + 2 \times 0.75\)

\(a_3 = 60 + 1.5\)

\(a_3 = 61.50\)