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# An auditorium with 30 rows of seats has 10 seats in the first row. Each successive row has one more seat than the previous row. If students

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An auditorium with 30 rows of seats has 10 seats in the first row. Each successive row has one more seat than the previous row. If students taking an exam are permitted to sit in any row, but not next to another student in that row, what is the maximum number of students that can be seated for an exam?

Aug 20, 2017

#1
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First, the number of seats in the 30 rows form an arithmetic series as follows:

10, 11, 12.........etc.

Sum =n/2[2f + (n - 1)d

Sum = 30/2[2*10 + (30 - 1)*1

Sum = 15 [20 + 29]

Sum = 15 x 49

Sum = 735 total number of seats in the auditorium.

Since the students have to be separated by at least one seat, the total number of students should be:

735 / 2 = 368 (rounded), maximum number of students.

If the seats were numbered from 1 to 735, then the students could be seated in seats:

1, 3, 5, 7......735, which would be a total of 368 students, which includes the first student in seat No. 1 and the last student in seat No. 735.

Aug 21, 2017
#2
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An auditorium with 30 rows of seats has 10 seats in the first row. Each successive row has one more seat than the previous row. If students taking an exam are permitted to sit in any row, but not next to another student in that row, what is the maximum number of students that can be seated for an exam?

Nice effort guest, you just messed up a little with how many students can be seated in each odd row.

How many seats are there in the audirtorium. I do not need to find this but I want to check my answer above the one above.  S30=30/2(20+29*1)=15*49=735       A match

5 people first row  out of 10 seats

6 people second row out of 11 seats

6 people second row out of 12 seats

7 people second row out of 13 seats

7 people second row out of 14 seats

....

How many seats are in the 30th row?

\(T_{30}=10+29*1=39seats\)

20 people 30th row out of 39 seats

Number of people

= (5+6+.....+19)+(6+7+...+20)

= 5+2(6+7+...+19)+20

=25    +   2[n/2(a+L)]

=25    +    2[ 14/2(6+19)]

=25    +     14*(25)

=25    +     350

=375

A total of 375 students can be seated in the auditorium :)

Aug 22, 2017