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?
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
What formula did you use? What is it called?
I learned na bogo jud kog math
Way pulos
To find the number of rows of seats in the auditorium, we need to determine the value of the last row.
We know that the first row has 21 seats, and each subsequent row has one more seat than the row in front of it.
We can use the formula for the sum of an arithmetic series to find the number of seats in each row.
The sum formula is given by: Sn = (n/2)(a+L), where Sn is the sum of the series, n is the number of terms, a is the first term, and L is the last term.
In this case, we know that a = 21 seats and Sn = 1065 seats. We need to find the value of L.
Substituting these values into the formula, we have: 1065 = (n/2)(21 + L).
Simplifying the equation, we get: 1065 = (21/2)(2 + L).
Multiplying both sides by 2 to eliminate the fraction, we have: 2130 = 21(2 + L).
Expanding the equation, we get: 2130 = 42 + 21L.
Subtracting 42 from both sides, we have: 2088 = 21L.
Dividing both sides by 21, we get: L = 99.
Therefore, the last row has 99 seats.
Now, we can find the number of rows in the auditorium by summing the first and last rows.
Dividing the total number of seats by the average number of seats per row (the sum of the first and last rows divided by 2), we have:
Number of rows = 1065 / ((21 + 99) / 2).
Calculating the average number of seats per row, we have:
Number of rows = 1065 / (120 / 2).
Number of rows = 1065 / 60.
Number of rows ≈ 17.75.
Since the number of rows must be a whole number, we need to round up to the nearest integer.
Therefore, there are approximately 18 rows of seats in the auditorium.