1. **Find the 15th term of the arithmetic sequence: 3, 7, 11, ...** The formula for the $n$th term of an arithmetic sequence is: $$a_n = a_1 + (n-1)d$$ where $a_1$ is the first term and $d$ is the common difference. Here, $a_1 = 3$ and $d = 7 - 3 = 4$. Calculate the 15th term: $$a_{15} = 3 + (15-1) \times 4 = 3 + 14 \times 4 = 3 + 56 = 59$$ 2. **Determine the common difference and first term if $a_{10} = 52$, $a_{20} = 102$** Using the formula: $$a_n = a_1 + (n-1)d$$ We have two equations: $$a_{10} = a_1 + 9d = 52$$ $$a_{20} = a_1 + 19d = 102$$ Subtract the first from the second: $$a_{20} - a_{10} = (a_1 + 19d) - (a_1 + 9d) = 10d = 102 - 52 = 50$$ So, $$d = \frac{50}{10} = 5$$ Substitute back to find $a_1$: $$a_1 + 9 \times 5 = 52 \Rightarrow a_1 + 45 = 52 \Rightarrow a_1 = 7$$ 3. **Write the explicit formula for the sequence: 5, 9, 13, ...** First term $a_1 = 5$, common difference $d = 9 - 5 = 4$. Formula: $$a_n = 5 + (n-1) \times 4 = 5 + 4n - 4 = 4n + 1$$ 4. **Find $n$ if the $n$th term is 101 in the sequence: 1, 5, 9, ...** First term $a_1 = 1$, common difference $d = 5 - 1 = 4$. Set $a_n = 101$: $$101 = 1 + (n-1) \times 4$$ Simplify: $$101 - 1 = 4(n-1) \Rightarrow 100 = 4(n-1)$$ Divide both sides by 4: $$\cancel{4}(n-1) = \frac{100}{\cancel{4}} = 25$$ So, $$n - 1 = 25 \Rightarrow n = 26$$ **Final answer for the first problem:** $$\boxed{59}$$