Since this is an arithmetic progression, the sequence is defined recursively by
\(a_n=a_{n-1}+d\)
where \(d\) is the common difference and \(a_n\) is the \(n\)th term. Solving for \(a_n\) gives
\(a_n=a_{n-1}+d=a_{n-2}+2d+a_{n-3}+3d=\cdots=a_1+(n-1)d\)
Adding up the first 16 terms gives
\(\displaystyle\sum_{n=1}^{16}(a_1+(n-1)d)=16a_1+120d\)
Adding up the next 16 terms gives
\(\displaystyle\sum_{n=17}^{32}(a_1+(n-1)d)=16a_1+376d\)
So you have two equations with two unknowns,
\(\begin{cases}16a_1+120d=528\\16a_1+376d=1552\end{cases}\)
Solving gives \(a_1=3\) and \(d=4\).
