well, the sequence goes, 1100, to 1135, to 1170.... notice, is simply adding 35 to get the next term, so the common difference is 35, and the first term is 1100 of course.
[tex]\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\[-0.5em] \hrulefill\\ a_1=1100\\ d=35\\ a_n=3725 \end{cases} \\\\\\ 3725=1100+(n-1)35\implies 3725=1100+35n-35 \\\\\\ 3725=1065+35n\implies 2660=35n\implies \cfrac{2660}{35}=n\implies 76=n[/tex]
