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