Question 178126
all the even rows contain 1/2 the number of students.
there are 10 even rows, each containing half the number of students as seats.
you start with 10 seats which means the first even row contains 5 students.
this winds up being an arithmetic series where n = 10.
you don't start off with 1, however.
you want to start with 5.
this means that you will need to subtract the sum of the first 4 numbers.
4 + 10 = 14
you take the sum of 1 through 14 and you subtract the sum of 1 through 4 and you will get the sum of 5 through 14 as your answer.
s = n*(n+1)/2
when n = 14, you get:
s = 14*15/2 = 7*15 = 105
when n = 4, s = 4*5/2 = 10
105-10 = 95
---
you have 10 odd rows.
each odd row contains 1/2 the number of students truncated to the nearest integer + 1.
as an example:
11 students = 11/2 = 5.5 students = 5 + 1 = 6
---
the first odd row contains 6.
each succeeding odd row contains 1 more.
the odd rows are also an arithmetic series.
you are going to start with 6 so you would need to subtract the first 5.
5 + 10 = 15 total
---
when n = 15, you get:
s = 15*16/2 = 15*8 = 120
when n = 5, you get:
s = 5*6/2 = 15
120 - 15 = 105
---
the sum of the number of students in the even and odd rows should be
95 + 105 = 200
---
your answer is:
maximum number of students that can be seated is 200.
---
here's an example of an even row with 10 seats.
x is the student, . is the space.
x.x.x.x.x.
---
here's an example of an odd row with 11 seats.
x.x.x.x.x.x
---
here's what your total seating arrangement will look like, starting with 10 seats in the first row and adding one seat as you go down.
01-x.x.x.x.x. = 5
02-x.x.x.x.x.x = 6
03-x.x.x.x.x.x. = 6
04-x.x.x.x.x.x.x = 7
05-x.x.x.x.x.x.x. = 7
06-x.x.x.x.x.x.x.x = 8
07-x.x.x.x.x.x.x.x. = 8
08-x.x.x.x.x.x.x.x.x = 9
09-x.x.x.x.x.x.x.x.x. = 9
10-x.x.x.x.x.x.x.x.x.x = 10
11-x.x.x.x.x.x.x.x.x.x. = 10
12-x.x.x.x.x.x.x.x.x.x.x = 11
13-x.x.x.x.x.x.x.x.x.x.x. = 11
14-x.x.x.x.x.x.x.x.x.x.x.x = 12
15-x.x.x.x.x.x.x.x.x.x.x.x. = 12
16-x.x.x.x.x.x.x.x.x.x.x.x.x = 13
17-x.x.x.x.x.x.x.x.x.x.x.x.x. = 13
18-x.x.x.x.x.x.x.x.x.x.x.x.x.x = 14
19-x.x.x.x.x.x.x.x.x.x.x.x.x.x. = 14
20-x.x.x.x.x.x.x.x.x.x.x.x.x.x.x = 15
---
the odd numbered rows have an even number of seats.
the even numbered rows have an odd number of seats.
---
sum of number is 1*5 + 2*(2 thoruh 14) + 1*15
sum is:
5
12
14
16
18
20
22
24
26
28
15
total is: 200
---
this agrees with the arithmetic sequence logic which i suspect is where they were driving you.
---