Is it an AP?

$a_n = 5n - 7$ .
Replacing $n$ with $n + 1$ , we get the ${(n + 1)}^{\text{th}}$ term.
$a_{n + 1} = 5\left(n + 1\right) - 7\\ a_{n + 1} = 5n + 5 - 7\\ a_{n + 1} = 5n - 2$

The common difference(d) between the terms of this AP can be found by subtracting the ${(n + 1)}^{\text{th}}$ term from the $n^{\text{th}}$ term.

So, $a_{n + 1} - a_n = \left(5n - 2\right) - \left(5n - 7\right)\\ d = 5n - 2 - 5n + 7\\ d = 5$

So, the common difference is a constant term and independent of $n$ . So, $\color{#69047E}{\boxed{\text{Yes}}}$ , this sequence is an AP.

Generalizing:-
Let the $n^{\text{th}}$ term of a sequnce be defined by a linear equation $xn + y$ where $x$ and $y$ are constants.
$a_n = xn + y$ .
Replacing $n$ with $n + 1$ , we get the ${(n + 1)}^{\text{th}}$ term.
$a_{n + 1} = x\left(n + 1\right) + y\\ a_{n + 1} = xn + x + y$

The common difference(d) between the terms of this AP can be found by subtracting the ${(n + 1)}^{\text{th}}$ term from the $n^{\text{th}}$ term.

So, $a_{n + 1} - a_n = \left(xn + y\right) - \left(xn + x + y\right)\\ d = xn + x + y - xn - y\\ d = x$

So, the common difference is a constant $x$ . So, any sequence whose $n^{\text{th}}$ term can be defined as a linear expression in $n$ , is in arithmetic progression and the common difference between its terms is the coefficient of $n$ .

Just put $n$ as 1,2,3.And see the magic. ;)