The 19th term of an A.P. is equal to three times its 6th term. If its 9th term is 19, find the A.P.

We know that a n = a + (n - 1)d According to question, ⇒ a 19 = 3(a 6 ) ⇒ a + 18d = 3(a + 5d) ⇒ a + 18d = 3a + 15d ⇒ 3a - a = 18d - 15d ⇒ 2a = 3d ⇒ a = (Eq 1) Given, a 9 = 19 Putting value of d in Eq 1 we get, Hence, the terms of A.P. are a 2 = a 1 + d = 3 + 2 = 5, a 3 = a 2 + d = 5 + 2 = 7, a 4 = a 3 + d = 7 + 2 = 9. Hence, the terms of the A.P. are 3, 5, 7, 9, ....