Asked by Roxane
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
Answered by
Reiny
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
Answered by
Yawa mo
Bubu kayu
Answered by
chad
What formula did you use? What is it called?
Answered by
Litsi
Way pulos
Answered by
30 rows
I learned na bogo jud kog math
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.