Answer: 3, 7, 11, . . . , 147
The middle term of 37 terms is \(t_{19}\). The middle three terms are \(t_{18}, t_{19}, t_{20}\):
\[(a + 17d) + (a + 18d) + (a + 19d) = 225\]\[a + 18d = 75 \quad \text{(I)}\]Last three terms \(t_{35}, t_{36}, t_{37}\):
\[3a + 105d = 429\]\[a + 35d = 143 \quad \text{(II)}\](II) − (I): \(17d = 68\), \(d = 4\); \(a = 75 - 72 = 3\). Last term \(t_{37} = 3 + 36(4) = 147\).