Answer:
D) 35, 38, 41, 44, 47, 50, ...
Step-by-step explanation:
tₙ = 32 + 3n; n ∈ N, n ≥ 1
t₁ = 32 + 3(1) = 32 + 3 = 35
t₂ = 32 + 3(2) = 32 + 6 = 38
t₃ = 32 + 3(3) = 32 + 9 = 41
t₄ = 32 + 3(4) = 32 + 12 = 44
t₅ = 32 + 3(5) = 32 + 15 = 47
t₆ = 32 + 3(6) = 32 + 18 = 50
The sequence starts at t₁ = 35 and each term is 3 units greater than the preceding term.
The sequence is 35, 38, 41, 44, 47, 50, ...