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.
2007-07-25 06:14:35 · 9 answers · asked by jessie0420 2 in Science & Mathematics ➔ Mathematics
131 seats in 38th row. so over 2000 seats is correct.
2007-07-25 06:22:53 · answer #1 · answered by michelle 2 · 0⤊ 0⤋
By observation, we establish that the number of seats on each row increases by 3 from the total number of seats in the prior row.
The equation -based on the above observed behavior - that will compute the number of seats in an xth row will therefore be:
Nx = 20 + 3(x -1)
Rationale for the formula - We have a base number of seats of 20 in row one. Any increment of 3 is a function of how far is the subject row X from the first row. Thus we deduct 1. For example, row 2 is one removed from the base row 1 will have
an increment of 3 in seats;
Proof; N2 = 20 + 3(2-1) = 20 + 3 (1) = 23. That's easy because we visualize there is only one row difference, and therefore, an added 3 searts only to row 1.
Now to solve for the number of seats on Row 38, using the equation we established:
N38 = 20 + 3(38 -1)
N38 = 20 + 3 (37)
N38 = 20 + 111
N38 = 131 seats.
I hope my solution "sits" well with you.
Watch the games and be comfortable with your seat!
2007-07-25 13:31:17 · answer #2 · answered by the lion and the bee 3 · 0⤊ 0⤋
It is an arithmetic progression,
Final term(seats in the last row) = A1 + (n - 1)d
A1 = 20, n = 38, d = 3
Final term = 20 + (38 - 1)(3)
Final term = An = 131 seats.
SUM = n(A1 + An)/2
SUM = 38/2 (20 + 131)
SUM = 19 (151)
SUM = 2869
sum = total no. of seats = 2869.
2007-07-25 13:22:35 · answer #3 · answered by Anonymous · 1⤊ 0⤋
Arithmetic series:
Sn = (n /2) [ 2a + (n - 1) d) ]
a = 20
n = 38
d = 3
Sn = 19 [ 40 + (37 x 3)]
Sn = 19 x 151
Sn = 2869
There are 2869 seats in this part of the stadium.
2007-07-25 14:58:19 · answer #4 · answered by Como 7 · 0⤊ 0⤋
Answer: Total seats in section =2869.
Sigma(20+3*k, k = 0..37)
2007-07-25 13:41:05 · answer #5 · answered by ? 5 · 0⤊ 0⤋
First take the easy part, the fixed number of seats
20*38=76
The next part is adding up the extra seats
3*(1+2+...+37)
(1+2+3+..+26+37)=37*38/2
The rest is pretty easy.
2007-07-25 13:29:34 · answer #6 · answered by 2 meter man 3 · 0⤊ 0⤋
131 Seats
2007-07-25 13:19:54 · answer #7 · answered by Michael S 2 · 0⤊ 0⤋
37
â (20 + 3n) = 20 + (20 + 3) + (20 + 6) + ...+ (20 + 3(37)) =
n = 0
20(38) + 3[37(38) / 2] = 760 + 2109 = 2869.
2007-07-25 13:36:57 · answer #8 · answered by S. B. 6 · 0⤊ 0⤋
it is A.S. a= 20 , d= 3 n = 38
sum =n/2( 2a +(n-1)d) = 38/2(2x20 + 37x3)=2869
2007-07-25 14:36:19 · answer #9 · answered by mramahmedmram 3 · 0⤊ 0⤋