If the series is made up of numbers in an arithmetic progression, you have
\(\displaystyle\sum_{n=1}^{18}a_n=\sum_{n=1}^{18}(a_1+(n-1)d)=18a_1+153d=4185\)
where \(a_1\) is the first term and \(d\) is the common difference between terms.
Since the last term in the series is \(a_{18}=275\), you have
\(a_{18}=a_1+(18-1)d\implies a_1+17d=275\)
Solve the system
\(\begin{cases}18a_1+153d=4185\\a_1+17d=275\end{cases}\)
and you'll find that the first term is \(a_1=190\) (with \(d=5\)).
