Arithmetic progression 1

$S_n = pn+qn^2\\ T_1 = S_1 = p(1) + q(1)^2 = p+q\\ T_2 = S_2 - S_1\\ =p(2) + q(2^2) - (p+q)\\ =2p+4q -p-q\\ =p+3q\\ d= T_2 - T_1\\ =p+3q-(p+q)\\ =\boxed{2q}$

There is a simple approach to this problem.
Now $S_n = pn + qn^2$ .
If we replace $n$ with $n - 1$ it gives $S_{n - 1}$ .
So, $S_{n - 1} = p(n - 1) + q{(n - 1)}^2\\ S_{n - 1} = pn - p + q\left(n^2 - 2n + 1\right)\\ S_{n - 1} = pn - p + qn^2 - 2qn + q$

Now if we subtract $S_{n - 1}$ from $S_n$ we get the nth term $a_n$ .
$a_n = S_n - S_{n - 1}\\ a_n = pn + qn^2 - \left(pn - p + qn^2 - 2qn + q\right)\\ a_n = pn + qn^2 - pn + p - qn^2 + 2qn - q\\ a_n = p + q(2n - 1)$

Here we see that any term of an is defined. So, if we replace $n$ with $n - 1$ we get $a_{n - 1}$ .
$a_{n - 1} = p + q\left(2(n - 1) - 1\right)\\ a_{n - 1} = p + q\left(2n - 3\right)$

Since $a_{n - 1}$ and $a_n$ are consecutive terms of an AP their difference is the common difference of the AP.
$a_n - a_{n - 1} = d\\ d = p + q(2n - 1) - \left(p + q(2n - 3)\right)\\ d = p - p + q\left(2n - 1 - 2n + 3\right)\\ d = \color{#69047E}{\boxed{2q}}$