An arithmetic sequence has 2036 terms labelled $t_{1}, t_{2}, t_{3}, \ldots, t_{2035}, t_{2036}$. Its 2018 th term is $t_{2018}=100$. Determine the value of $t_{2000}+5 t_{2015}+5 t_{2021}+t_{2036}$.

(An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant, called the common difference. For example, 3, 5, 7, 9 are the first four terms of an arithmetic sequence with common difference 2.)
Suppose that $d$ is the common difference in this arithmetic sequence.

Since $t_{2018}=100$ and $t_{2021}$ is 3 terms further along in the sequence, then $t_{2021}=100+3 d$.

Similarly, $t_{2036}=100+18 d$ since it is 18 terms further along.

Since $t_{2018}=100$ and $t_{2015}$ is 3 terms back in the sequence, then $t_{2015}=100-3 d$.

Similarly, $t_{2000}=100-18 d$ since it is 18 terms back.

Therefore,

$$
\begin{aligned}
t_{2000}+5 t_{2015}+5 t_{2021}+t_{2036} & =(100-18 d)+5(100-3 d)+5(100+3 d)+(100+18 d) \\
& =1200-18 d-15 d+15 d+18 d \\
& =1200
\end{aligned}
$$

ANSWER: 1200