1. **Problem Statement:** We need to find the total seating capacity of a theatre with 40 rows of chairs. 2. **Given:** - First row has 10 chairs. - Each subsequent row has 2 more chairs than the previous row. - Total rows = 40. 3. **Formula:** This is an arithmetic sequence where the first term $a_1 = 10$, common difference $d = 2$, and number of terms $n = 40$. The number of chairs in the $n^{th}$ row is given by: $$a_n = a_1 + (n-1)d$$ The total seating capacity is the sum of all chairs in 40 rows, which is the sum of an arithmetic series: $$S_n = \frac{n}{2} (a_1 + a_n)$$ 4. **Calculate the number of chairs in the 40th row:** $$a_{40} = 10 + (40-1) \times 2 = 10 + 39 \times 2 = 10 + 78 = 88$$ 5. **Calculate the total seating capacity:** $$S_{40} = \frac{40}{2} (10 + 88) = 20 \times 98 = 1960$$ 6. **Answer:** The theatre has a total seating capacity of **1960 chairs**.