Solution: Let the arithmetic sequence have first term $ a_1 = 15 $ and fifth term $ a_5 = 35 $. The common difference $ d $ satisfies $ a_5 = a_1 + 4d $. Substituting: $ 35 = 15 + 4d \Rightarrow d = 5 $. The sum of the first $ n = 10 $ terms is $ S_{10} = rac{10}{2} [2(15) + 9(5)] = 5(30 + 45) = 5(75) = 375 $. oxed{375}

["Solution: Arithmetic Sequence with First Term 15 and Fifth Term 35", "Understanding arithmetic sequences is fundamental in mathematics, especially in algebra and sequences. In this problem, we analyze a sequence where the first term is $ a_1 = 15 $ and the fifth term is $ a_5 = 35 $. By leveraging the explicit formula for any term in an arithmetic sequence, we solve for the common difference and compute the sum of the first 10 terms.", "### Step 1: Recall the formula for the $ n $-th term", "In an arithmetic sequence, each term is given by:\n$$\na_n = a_1 + (n - 1)d\n$$\nwhere $ a_1 $ is the first term, $ d $ is the common difference, and $ n $ is the term number.", "Given:\n- $ a_1 = 15 $\n- $ a_5 = 35 $", "Using the formula for the fifth term:\n$$\na_5 = a_1 + 4d\n$$\nSubstitute known values:\n$$\n35 = 15 + 4d\n$$\nSolving for $ d $:\n$$\n4d = 35 - 15 = 20 \quad \Rightarrow \quad d = \frac{20}{4} = 5\n$$", "### Step 2: Use the sum formula for the first $ n $ terms", "The sum of the first $ n $ terms of an arithmetic sequence is given by:\n$$\nS_n = \frac{n}{2} \left[ 2a_1 + (n - 1)d \right]\n$$\nAlternatively, it can also be computed using:\n$$\nS_n = \frac{n}{2} (a_1 + a_n)\n$$\nSince $ n = 10 $, we compute $ a_{10} $ using the common difference $ d = 5 $:\n$$\na_{10} = a_1 + 9d = 15 + 9 \ imes 5 = 15 + 45 = 60\n$$\nNow apply the sum formula:\n$$\nS_{10} = \frac{10}{2} (a_1 + a_{10}) = 5 \ imes (15 + 60) = 5 \ imes 75 = 375\n$$", "### Conclusion", "This arithmetic sequence has a first term of 15 and common difference 5, resulting in a total sum of 375 for the first 10 terms. Mastering term and sum formulas enables efficient problem-solving in sequences and series.", "\boxed{375}"]