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?
2 answers
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
a+(n-1)d = term(n)
18 + n-1 = 51
n = 51+1-18 = 34
34 rows