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,
a5 + a9 = 30
⇒ a + (5 − 1)d + a + (9 − 1)d = 30
⇒ a + 4d + a + 8d = 30
⇒ 2a + 12d = 30
⇒ a + 6d = 15 .... (1)
Also, a25 = 3(a8)
⇒ a + (25 − 1)d = 3[a + (8 − 1)d]
⇒ a + 24d = 3a + 21d
⇒ 3a − a = 24d − 21d
⇒ 2a = 3d
⇒ a = d ....(2)
Substituting the value of (2) in (1), we get
d + 6d = 15
⇒ 3d + 12d = 15 × 2
⇒ 15d = 30
⇒ d = 2
⇒ a = [From (1)]
⇒ a = 3
Thus, the A.P. is 3, 5, 7, 9, .... .