1. **Problem statement:** Given an arithmetic sequence where the 9th term $U_9 = N + 8$ and the sum of the first 9 terms $S_9 = 9 \times (N + 2)$, with $N=42$ (last two digits of NIM). 2. **Known formulas and rules:** - The $n$th term of an arithmetic sequence: $$U_n = a + (n-1)d$$ where $a$ is the first term and $d$ is the common difference. - The sum of the first $n$ terms: $$S_n = \frac{n}{2} (2a + (n-1)d)$$ 3. **Substitute given values:** - $U_9 = 42 + 8 = 50$ - $S_9 = 9 \times (42 + 2) = 9 \times 44 = 396$ 4. **Express $U_9$ and $S_9$ in terms of $a$ and $d$:** - $$U_9 = a + 8d = 50$$ - $$S_9 = \frac{9}{2} (2a + 8d) = 396$$ 5. **Simplify $S_9$ equation:** - Multiply both sides by 2: $$9 (2a + 8d) = 792$$ - Divide both sides by 9: $$2a + 8d = \cancel{\frac{792}{9}}88$$ 6. **Rewrite the system:** - From $U_9$: $$a + 8d = 50$$ - From $S_9$: $$2a + 8d = 88$$ 7. **Subtract the first equation from the second:** - $$ (2a + 8d) - (a + 8d) = 88 - 50$$ - $$a = 38$$ 8. **Find $d$ using $a + 8d = 50$:** - $$38 + 8d = 50$$ - $$8d = 12$$ - $$d = \frac{12}{8} = \frac{3}{2} = 1.5$$ 9. **Calculate $U_5$:** - $$U_5 = a + 4d = 38 + 4 \times 1.5 = 38 + 6 = 44$$ 10. **Calculate $S_5$:** - $$S_5 = \frac{5}{2} (2a + 4d) = \frac{5}{2} (2 \times 38 + 4 \times 1.5) = \frac{5}{2} (76 + 6) = \frac{5}{2} \times 82 = 5 \times 41 = 205$$ 11. **Answer part (a):** - $$U_5 - 2 = 44 - 2 = 42$$ 12. **Answer part (b):** - Calculate $S_9 - S_5 - 23$: $$396 - 205 - 23 = 168$$ - Divide by 4: $$\frac{168}{4} = 42$$ **Final answers:** - (a) $U_5 - 2 = 42$ - (b) $\frac{1}{4} (S_9 - S_5 - 23) = 42$