This problem follows an arithmetic sequence, where:

• The first term $a_1 = 24$ (seats in the first row).
• The common difference $d = 2$ (each row has 2 more seats than the previous row).
• The number of terms $n = 20$ (total rows in the auditorium).

The formula for the sum of an arithmetic sequence is:

$S_n = \frac{n}{2} \times (2a_1 + (n-1)d)$

Step 1: Substitute the given values

$S_{20} = \frac{20}{2} \times (2(24) + (20-1)(2))$

$S_{20} = 10 \times (48 + 38)$

$S_{20} = 10 \times 86$

$S_{20} = 860$

Final Answer:

The auditorium requires 860 steel seats.

Would you like a breakdown of any step? Here are some related questions you might find interesting:

1. How many seats are in the last row?
2. What if the common difference was 3 instead of 2?
3. How would the total change if there were 25 rows instead?
4. What is the total number of seats if each row had 30 seats instead of 24 initially?
5. What is the total number of seats if the pattern continued for 30 rows?

Tip: Arithmetic sequences are useful in real-life scenarios like seating arrangements, staircase steps, and salary increments!