The general term of an arithmetic sequence is tn = 32 + 3n, where n ∈n and n ≥ 1. the sequence is
a) 32, 35, 38, 41, 44,
b) 32, 96, 288, 864, 2592,
c) 34, 39, 44, 49, 54,
d) 35, 38, 41, 44, 47,

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, ...

step-by-step explanation:

try remembering the steps

well what you have to do is divide 84 by 6 so theanswer

step-by-step explanation:

84 divided by 6 = 14

and that lightbulb changing party must be freaking lit!