Question
The total seating capacity of an auditorium is 1,065. The first row has 21 seats arithmetic series and each row has one seat more than the row in front of it. How many rows of seats are there in the auditorium?
Answers
In your arithmetic series:
a = 21
d = 1
n = ??
sum(n) = 1065
1065 = (n/2)(2(21) + (n-1)(1) )
2130 = n(42 + n - 1)
2130 = n^2 + 41n
n^2 + 41n - 2130 = 0
(n-30)(n + 71) = 0
n = 30 or n = -71 , but n > 0
there are 30 rows
a = 21
d = 1
n = ??
sum(n) = 1065
1065 = (n/2)(2(21) + (n-1)(1) )
2130 = n(42 + n - 1)
2130 = n^2 + 41n
n^2 + 41n - 2130 = 0
(n-30)(n + 71) = 0
n = 30 or n = -71 , but n > 0
there are 30 rows
Bubu kayu
What formula did you use? What is it called?
Way pulos
I learned na bogo jud kog math
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