A section in a stadium has 20 seats in the first row, 23 seats in the second row, increasing by 3 seats each row for a total of 38 rows. How many seats are in this section of the stadium?
Please explain using words, numbers, and formulas.
Recall that Sn = n/2 (2a+(n-1)d)
2869 seats
Heh 69
Well, this section of the stadium has a total of 38 rows. In the first row, there are 20 seats. In the second row, there are 23 seats. And for each subsequent row, the number of seats increases by 3.
To find out how many seats are in this section of the stadium, we can first calculate the number of seats in each row using the formula:
Number of seats in a row = 20 + (row number - 1) * 3
Let's calculate the number of seats for each row:
Row 1: 20 + (1-1)*3 = 20
Row 2: 20 + (2-1)*3 = 23
Row 3: 20 + (3-1)*3 = 26
...
Row 38: 20 + (38-1)*3 = 131
Now, to find the total number of seats in the section, we can add up the number of seats in each row:
Total seats = Sum of seats in each row
= 20 + 23 + 26 + ... + 131
To calculate this sum, we can use the formula for the sum of an arithmetic series:
Sum = (first term + last term) * number of terms / 2
In this case, the first term is 20, the last term is 131, and the number of terms is 38:
Total seats = (20 + 131) * 38 / 2
= 151 * 38 / 2
= 5749
So, there are a total of 5749 seats in this section of the stadium.
To find the total number of seats in this section of the stadium, we need to calculate the sum of the number of seats in each row.
The first row has 20 seats.
The second row has 23 seats.
The number of seats in each subsequent row increases by 3.
To find the number of seats in the third row, we add 3 seats to the number of seats in the second row: 23 + 3 = 26 seats.
Similarly, to find the number of seats in the fourth row, we add another 3 seats: 26 + 3 = 29 seats.
We can notice that the number of seats in each row forms an arithmetic sequence, where the first term (a) is 20, the common difference (d) is 3, and the number of terms (n) is 38 rows.
Using the formula for the sum of an arithmetic sequence:
Sn = (n/2) * (2a + (n - 1)d),
where Sn is the sum of the sequence, we can substitute in the values we have:
S38 = (38/2) * (2 * 20 + (38 - 1) * 3) = 19 * (40 + 37 * 3) = 19 * (40 + 111) = 19 * 151 = 2879.
Therefore, there are 2,879 seats in this section of the stadium.
To find the number of seats in this section of the stadium, we need to calculate the sum of the seats in each row. We know that the first row has 20 seats, and each subsequent row has 3 more seats than the previous row.
The general formula for calculating the sum of an arithmetic series is:
Sum = (n/2) * (first term + last term)
In this case, the first term is 20, and we need to find the last term. To find the last term, we can use the formula for the nth term of an arithmetic sequence:
nth term = first term + (n - 1) * common difference
The common difference in this case is 3, and we have a total of 38 rows. Plugging these values into the formula, we can find the last term:
38th term = 20 + (38 - 1) * 3 = 20 + 37 * 3 = 20 + 111 = 131.
Now we can calculate the sum of the seats in each row:
Sum = (38/2) * (20 + 131) = 19 * 151 = 2879.
Therefore, there are 2879 seats in this section of the stadium.