The 19th term of an A.P. is equal to three times its sixth term.

Let a be the first term and d be the common difference.

We know that, nth term = an = a + (n − 1)d

According to the question,

a19 = 3a6
⇒ a + (19 − 1)d =  3(a + (6 − 1)d)
⇒ a + 18d =  3a + 15d
⇒ 18d − 15d =  3a − a
⇒ 3d = 2a
⇒ a = 32$=\frac{3}{2} d$  .... (1)

Also, a9 = 19
⇒ a + (9 − 1)d = 19
⇒ a + 8d = 19   ....(2)

On substituting the values of (1) in (2), we get
32$\frac{3}{2} d$ + 8d = 19
⇒ 3d + 16d = 19 × 2
⇒ 19d = 38
⇒ d = 2
⇒ a = 32×2$=\frac{3}{2}$×2     [From (1)]
⇒ a = 3

Thus, the A.P. is 3, 5, 7, 9, .... .