The first four terms in a sequence of numbers are $3 + k,\ 1 + k,\ -1 + k,\ -3 + k,\ \dots$ where $k$ is a constant.

1. Identify the common difference and form the nth term.

The terms decrease by $2$ each time, so the common difference is $-2$.

Using $T_n=a+(n-1)d$ with $a=3+k$ and $d=-2$,
$T_n=3+k+(n-1)(-2)$
$=3+k-2n+2=5+k-2n$

1. Set the nth term equal to 39 and solve for $k$ in terms of $n$.

From part (a), the $n$th term is $5+k-2n$.

For 39 to be a term,
$5+k-2n=39$
$k=34+2n$

2. Convert the condition on $n$ into conditions on $k$.

Since $n$ is a positive integer, $2n$ is a positive even integer.

So $k=34+2n$ means $k$ must be even and greater than $34$.