The given Arithmetic Progression (A.P.) is \(3, 15, 27, 39, \dots\). The first term is \(a = 3\), and the common difference is \(d = a_2 - a_1 = 15 - 3 = 12\). The 54th term is calculated as \(a_{54} = a + (54 - 1) d = 3 + (53)(12) = 3 + 636 = 639\). We are looking for the term that is \(132\) more than the 54th term, which is \(639 + 132 = 771\). Let the \(n^{th}\) term be \(771\). Using the formula for the \(n^{th}\) term, \(a_n = a + (n - 1) d\), we have \(771 = 3 + (n - 1) 12\). This simplifies to \(768 = (n - 1) 12\), so \(n - 1 = \frac{768}{12} = 64\). Therefore, \(n = 65\).
Thus, the 65th term is \(132\) more than the 54th term.
Alternatively,
Let the \(n^{th}\) term be \(132\) more than the 54th term. The difference in terms is \(132\), and the common difference is \(12\). The number of additional common differences needed is \(\frac{132}{12} = 11\). Therefore, \(n = 54 + 11 = 65\). This means the 65th term is \(132\) more than the 54th term.