A60-row theater has 20 seats in the front row. the second row has 21 seats. if each row has one more than the row in front of it, how many seats are there in the theater?

The first row has 20 + 0 seats, the 2nd row, 20 + 1, 3rd row 20 + 2 . . . until 60th row has 20 + 59 (not 60 as the first row had 0 added on.

therefore you need to add up 20 + 21 + 22 . . . all the way to 76, 77, 78, 79.

79+ 21 = 100, 78 + 22 = 100, 77+ 23 = 100, in fact up to 49 + 51 = 100

That gives us 39 x 100 and add on the the other rows + 20 + 50

= 3970 seats

This is an arithmetic sequence with first term = 20 and common difference of 1 so we can use the sum formula :-

Sum of n terms ( Sn) = (n/2 )[2a1 + (n - 1) d]

S60 = (60/2) (2*20 + 59*1)

= 30 * 99

= 2970

4/6 = 2/3

so it should take 3/3 hours per man. so 9 hours is the answer 3*3 =9

step-by-step explanation:

this may not be at all, but i use the same website you're using for school, and if i'm stuck doing one of these i try to look for it on the section.