Answer
First sequence: $12,9,6,3,...$
$n$th term of the first sequence $a_{n}=-3n+15$
Second sequence: $4,1,-2,-5,...$
$n$th term of the second sequence $a_{n}=-3n+7$
Work Step by Step
For the first sequence,
The first term $a_{1}=12$ and
Common difference $d=-3$.
$n$th term of the sequence
$a_{n}=a_{1}+(n-1)d$
$=12+(n-1)(-3)$
$=12-3n+3=-3n+15$
For the second sequence,
The first term $a_{1}=4$ and
Common difference $d=-3$.
$a_{n}=4+(n-1)(-3)$
$=4-3n+3$
$=-3n+7$
