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

Given: a₂ + a₇ = 30

Also, a₁₅ = 2a₈ - 1

Consider a₂ + a₇ = 30

⇒ (a + d) + (a + 6d) = 30

⇒ 2a + 7d = 30 ……………….. (1)

Consider a₁₅ = 2a₈ - 1

⇒ a + 14d = 2(a + 7d) - 1

⇒ a + 14d = 2a + 14d - 1

⇒ a = 1

∴ First term = a = 1

Thus, from equation (1), we get,

7d = 30 - 2a

⇒ 7d = 30 - 2

⇒ 7d = 28

⇒ d = 4

Thus, the AP is a, a + d, a + 2d, a + 3d,…

Therefore, the AP is 1, 5, 9, 13, 17,…