so the first term is 36500.
after a year, he gets 2375 extra, so
1st year, 36500
2nd year 36500 + 2375
3rd year 36500 + 2375 + 2375
4th year 36500 + 2375 + 2375 + 2375
and so on
so the common difference is 2375, namely the number we add in order to get the following term.
[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=36500\\ d=2375 \end{cases} \\\\\\ a_n=36500+(n-1)2375\implies a_n=36500+2375(n-1) \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{his salary after 15 years, \boxed{n = 15}}}{a_{15}=36500+2375(15-1)}\implies a_{15}=36500+2375(14) \\\\\\ a_{15}=36500+33250\implies a_{15}=69750[/tex]
