Arithmetic Sequence D6F542

1. **Problem:** Find the 10th term of the arithmetic sequence where the first term $a_1=3$ and the common difference $d=5$. 2. **Formula:** The $n$th term of an arithmetic sequence is given by $$a_n = a_1 + (n-1)d$$ 3. **Step-by-step solution:** 1. Identify the given values: $a_1=3$, $d=5$, $n=10$. 2. Substitute into the formula: $$a_{10} = 3 + (10-1) \times 5$$ 3. Simplify inside the parentheses: $$a_{10} = 3 + 9 \times 5$$ 4. Multiply: $$a_{10} = 3 + 45$$ 5. Add: $$a_{10} = 48$$ 4. **Answer:** The 10th term is $48$. --- 1. **Problem:** Find the sum of the first 15 terms of an arithmetic sequence with first term $2$ and common difference $4$. 2. **Formula:** The sum of the first $n$ terms is $$S_n = \frac{n}{2} (2a_1 + (n-1)d)$$ 3. **Step-by-step solution:** 1. Given: $a_1=2$, $d=4$, $n=15$. 2. Substitute values: $$S_{15} = \frac{15}{2} (2 \times 2 + (15-1) \times 4)$$ 3. Simplify inside the parentheses: $$S_{15} = \frac{15}{2} (4 + 14 \times 4)$$ 4. Multiply: $$S_{15} = \frac{15}{2} (4 + 56)$$ 5. Add inside parentheses: $$S_{15} = \frac{15}{2} \times 60$$ 6. Multiply numerator: $$S_{15} = \frac{900}{2}$$ 7. Simplify fraction: $$S_{15} = 450$$ 4. **Answer:** The sum of the first 15 terms is $450$.