1. **Problem 1.1:** Find the sum of the first 15 terms of the arithmetic series 3 + 7 + 11 + 15 + ... 2. The formula for the sum of the first $n$ terms of an arithmetic series is: $$S_n = \frac{n}{2} (2a + (n-1)d)$$ where $a$ is the first term, $d$ is the common difference, and $n$ is the number of terms. 3. Identify the values: $a=3$, $d=7-3=4$, $n=15$. 4. Substitute into the formula: $$S_{15} = \frac{15}{2} (2 \times 3 + (15-1) \times 4)$$ 5. Simplify inside the parentheses: $$2 \times 3 = 6$$ $$(15-1) = 14$$ $$14 \times 4 = 56$$ 6. So: $$S_{15} = \frac{15}{2} (6 + 56) = \frac{15}{2} \times 62$$ 7. Multiply: $$S_{15} = \frac{15 \times 62}{2} = \frac{930}{2}$$ 8. Simplify the fraction: $$S_{15} = \cancel{\frac{930}{2}} = 465$$ --- 9. **Problem 1.2:** An arithmetic series has first term 8 and last term 92. The sum of all terms is 500. Find the number of terms. 10. The sum of an arithmetic series can also be calculated by: $$S_n = \frac{n}{2} (a + l)$$ where $l$ is the last term. 11. Given $a=8$, $l=92$, and $S_n=500$, substitute: $$500 = \frac{n}{2} (8 + 92)$$ 12. Simplify inside parentheses: $$8 + 92 = 100$$ 13. So: $$500 = \frac{n}{2} \times 100$$ 14. Multiply both sides by 2: $$1000 = n \times 100$$ 15. Divide both sides by 100: $$\cancel{\frac{1000}{100}} = n \cancel{\times \frac{100}{100}}$$ 16. Simplify: $$n = 10$$ --- 17. **Problem 1.3:** Find the 20th term of the series 2, 5, 8, 11, ... 18. The formula for the $n$th term of an arithmetic sequence is: $$a_n = a + (n-1)d$$ 19. Identify values: $a=2$, $d=5-2=3$, $n=20$. 20. Substitute: $$a_{20} = 2 + (20-1) \times 3$$ 21. Simplify: $$20-1=19$$ $$a_{20} = 2 + 19 \times 3 = 2 + 57 = 59$$ --- **Final answers:** 1.1 Sum of first 15 terms: $465$ 1.2 Number of terms: $10$ 1.3 20th term: $59$