3, 15, 27, 39, ………… an = ?, n = ?
an = a54 + 132
a = 3, d = 15 – 3 = 12
an = a54 + 132
an = a + 53d + 132
3 + 53(12) + 132
= 3 + 636 + 132
∴ an = 771
an = a + (n – 1) d = 771
= 3 + (n – 1)12 = 771
3 + 12n — 12 = 771
12n – 9 = 771
12n = 771 + 9
12n = 780
∴ n = \(\frac{780}{12}\)
∴ n = 65.
∴ 65th term is 132 more than its 54th term.