Answer:Step-by-step explanation:
The three arithmetic means between 10 and 34 are 16, 22, and 28.Problem 2: Insert four arithmetic means between 18 and 48Given:a_1 = 18a_6 = 48n = 6Solution:a_n = a_1 + (n - 1) da_48 = 18 + (6 - 1) d48 = 18 + 5d-5d = 18 - 48-5d/-5 = -30/-5d = 6Now that we know the common difference is 6, we can find the missing terms.First missing termsa_2 = 18 + da_2 = 18 + 6a_2 = 24Second missing terms a_3 = 24 + da_3 = 24 + 6a_3 = 30Third missing termsa_4 = 30 + da_4 = 30 + 6a_4 = 36Fourth missing terms a_5 = 36 + da_5 = 36 + 6a_5 = 42