Since this is an arithmetic progression, the sequence is defined recursively by
![a_n=a_{n-1}+d](/tpl/images/0211/8732/73d2d.png)
where
![d](/tpl/images/0211/8732/17a1d.png)
is the common difference and
![a_n](/tpl/images/0211/8732/3e3be.png)
is the
![n](/tpl/images/0211/8732/a4ae7.png)
th term. Solving for
![a_n](/tpl/images/0211/8732/3e3be.png)
gives
![a_n=a_{n-1}+d=a_{n-2}+2d+a_{n-3}+3d=\cdots=a_1+(n-1)d](/tpl/images/0211/8732/ad43e.png)
Adding up the first 16 terms gives
![\displaystyle\sum_{n=1}^{16}(a_1+(n-1)d)=16a_1+120d](/tpl/images/0211/8732/a76c1.png)
Adding up the next 16 terms gives
![\displaystyle\sum_{n=17}^{32}(a_1+(n-1)d)=16a_1+376d](/tpl/images/0211/8732/caacd.png)
So you have two equations with two unknowns,
![\begin{cases}16a_1+120d=528\\16a_1+376d=1552\end{cases}](/tpl/images/0211/8732/41291.png)
Solving gives
![a_1=3](/tpl/images/0211/8732/1c7e8.png)
and
![d=4](/tpl/images/0211/8732/160a9.png)
.