Good afternoon. How would a formula be made for this question? I believe it is arithmetic.
How many rows are in the corner section of a sports complex if the first row has 18 seats and the last row has 51 seats and each successice row has one additional seat?
r = 51 - 18
AS, with a = 18, d = 1, n = ? and term(n) = 51
a+(n-1)d = term(n)
18 + n-1 = 51
n = 51+1-18 = 34
34 rows
Good afternoon! To determine the number of rows in the corner section of a sports complex, we can use an arithmetic formula.
In this scenario, the first row has 18 seats and the last row has 51 seats, with each successive row having one additional seat.
To find the number of rows, we need to identify the pattern and use it to create our formula. We can observe that each row has one more seat than the previous row. So, the difference between the number of seats in two consecutive rows is 1.
We can express this pattern using an arithmetic sequence formula:
nth_term = first_term + (n - 1) * difference
Here, nth_term represents the number of seats in the nth row, first_term is the number of seats in the first row (18 in this case), n represents the position of the row, and difference is the difference in seats between consecutive rows (1 in this case).
To determine the number of rows in the corner section, we need to find the value of n when the number of seats in the nth row matches the last row (51 seats in this case).
Using the formula, we can substitute the given values:
51 = 18 + (n - 1) * 1
Now, we solve this equation for n:
51 - 18 = n - 1
33 = n - 1
n = 33 + 1
n = 34
So, there are 34 rows in the corner section of the sports complex.