Given ( a_{18} = 367 ), ( a_{30} = 499 )

[ a_n = a + (n-1)d ]

[ \begin{align*} a + 17d &= 367 \quad (1) \ a + 29d &= 499 \quad (2) \end{align*} ]

Subtract (1) from (2): [ 12d = 132 \implies d = 11 ]

From (1): [ a + 17\cdot 11 = 367 \implies a + 187 = 367 \implies a = 180 ]

We need ( a_n < 1000 ): [ 180 + (n-1)\cdot 11 < 1000 ] [ (n-1)\cdot 11 < 820 ] [ n-1 < \dfrac{820}{11} = 74.545\ldots ] [ n-1 \le 74 \implies n \le 75 ]

Answer: 75 terms