The given sequence is
12m, 15m, 18m, 21m, 24m, ...,
The 1st term is a₁ = 12m
The 2nd term is a₂ = 15m = a₁ + 3m = a₁ + (2-1)*3m
The 3rd term is a₃ = 18m = a₂ + 3m = a₁ + (3-1)*3m
Because each successive term is 3m more than the previous term, we have an arithmetic sequence with a common difference of 3.
Answer:
The n-th term is
[tex]a_{n} = a_{1} + (n-1)*(3m),\\for\, n=1,2,3,\,...,[/tex]
In recursive form,
[tex]a_{n+1}=a_{n}+3m[/tex]
