If each row have 6 more seats than the previous, then there is a common difference of 6 seats. This, in turn, becomes an arithmetic progression, with a₁ being 100 and d being 6. Now, we want to know the total number of seats when n = 23, so we need to use the sum of an arithmetic progression form:
[tex]S_n = \frac{n}{2}(2a_1 + (n - 1)d)[/tex]
[tex]S_{23} = \frac{23}{2}(2(100) + 22 \cdot 6)[/tex]
[tex]S_{23} = \frac{23}{2}(200 + 132)[/tex]
[tex]S_{23} = 3818[/tex]
Hence, there are 3818 seats in the first 23 rows.