1. **Stating the problem:** We have a sequence $(U_n)$ with the conditions: $$U_1 + U_2 + U_3 = 3$$ $$U_3 + U_4 + U_5 = 33$$ We need to find the value of $$U_{300} + U_{400} + U_{500}$$. 2. **Analyzing the problem:** The problem does not explicitly give a recurrence relation or formula for $(U_n)$, but since sums of three consecutive terms are given, it suggests the sequence might be arithmetic or follow a pattern. 3. **Assuming the sequence is arithmetic:** Let the first term be $a$ and common difference be $d$. Then: $$U_n = a + (n-1)d$$ 4. **Express the sums in terms of $a$ and $d$:** - Sum of first three terms: $$U_1 + U_2 + U_3 = a + (a+d) + (a+2d) = 3a + 3d = 3$$ - Sum of terms 3, 4, 5: $$U_3 + U_4 + U_5 = (a+2d) + (a+3d) + (a+4d) = 3a + 9d = 33$$ 5. **Solve the system:** From the first sum: $$3a + 3d = 3 \implies a + d = 1 \implies a = 1 - d$$ From the second sum: $$3a + 9d = 33$$ Substitute $a$: $$3(1 - d) + 9d = 33$$ $$3 - 3d + 9d = 33$$ $$3 + 6d = 33$$ $$6d = 30 \implies d = 5$$ Then: $$a = 1 - 5 = -4$$ 6. **Find $U_{300} + U_{400} + U_{500}$:** Calculate each term: $$U_{300} = a + 299d = -4 + 299 \times 5 = -4 + 1495 = 1491$$ $$U_{400} = a + 399d = -4 + 399 \times 5 = -4 + 1995 = 1991$$ $$U_{500} = a + 499d = -4 + 499 \times 5 = -4 + 2495 = 2491$$ Sum: $$1491 + 1991 + 2491 = 5973$$ **Final answer:** $$\boxed{5973}$$