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 = rac102 [2(15) + 9(5)] = 5(30 + 45) = 5(75) = 375 $. oxed375 - Nelissen Grade advocaten