Arithmetic and Geometric sequences - Describe the following sequence So I'm asked to determine the type of sequence below as well as state the $a$, $d$, and $r$ values. I know the answers to a, b, and c, but for the last one I'm confused as to what to categorize it under. I was thinking $a$ would be $65$, but that's all I could get. I can't figure out the d value, and I don't know whether it's geometric or arithmetic. Any help or hints would be appreciated.
$y = 77x - 12, x\in\mathbb{Z}^+$
I couldn't figure out how to format the equation properly, so I included a picture with the question below. I'm struggling with d)

As stated in the comments, a is the base value, d is the difference between each element in an arithmetic sequence, and r is the ratio between members in a geometric sequence. So like if I was given the pattern $3, 5, 7, \ldots$ 
$a = 3$, and $d = 2$. There is no $r$ value because it's an arithmetic sequence.
 A: The last equation (part d) means (though ambiguous) that you get the $n^{th}$ term of the sequence by plugging the values of $x$ in the R.H.S. For example - The first term of the sequence can be obtained by putting $x=1$ and it yields $(77*1-12=65)$. The $10^{th}$ term (for example) can be found by putting $x=10$ which yields $(77*10-12=758)$.
Now it is given that the domain of $x$ is the set of all positive integers i.e. $1,2,3...$ .
So the first term is $65$ (put $x=1$).
Second term is $142$ (put $x=2$).
Third term is $219$ (put $x=3$).
So it is clear that the sequence is an Arithmetic Progression with common difference  $77$.
Note - a, r and d are not fixed notations. You can rather say 
a = first term of the sequence,
d = common difference of an Arithmetic Progression,
r = $r^{th}$ term of the sequence.
A: COMMENT.- Question $d$ seems to me somewhat ambiguous. One way to interpret it is, for example, to take a positive integer arbitrary $x_0$ and define the sequence
$$x_1=77x_0-12\\x_2=77x_1-12\\x_3=77x_2-12$$ and so on so we have $$x_n=77^nx_0-12(77^{n-1}+77^{n-2}+\cdots+77+1)\\x_n=77x_0-12\frac{77^n-1}{77-1}=\frac{77^n(19x_0-3)+3}{19}$$ where $19x_0-3=a\in\mathbb N$ so $$x_n=\frac{77^na+3}{19}$$ However that is an interpretation of a question somewhat ambiguous as I said above.....
