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

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step-by-step explanation:

to find the total number of combinations, multiply the quantity of each item by each other:

2 pairs of shoes x 3 pairs of shorts = 6 different combinations.