An arithmetic sequence has first term 3 and common difference 7. From which term onwards are all terms greater than 2018?

Step-by-step solution

1. 1

   Write the general term. $u_n=3+(n-1)\cdot 7=7n-4$ The sequence is increasing, so once a term passes $2018$ all later terms do too.

2. 2

   Set up the strict inequality. $7n-4>2018\quad\Longrightarrow\quad 7n>2022\quad\Longrightarrow\quad n>\frac{2022}{7}\approx 288.86$

3. 3

   Round up to the next integer. The smallest integer strictly greater than $288.86$ is $n=289$.

4. 4

   Test the boundary terms explicitly. $u_{288}=7(288)-4=2012\le 2018,\qquad u_{289}=7(289)-4=2019>2018$

5. 5

   Answer. From the 289th term onwards every term exceeds $2018$, which is option B. Option C fails because $u_{288}=2012$ is still too small.