Question 1196225
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We have an arithmetic sequence because of the gap of 3 seats per row. 


Counting down the sequence involves -3, so going up the sequence involves +3 or 3.
a = first term
d = 3 = common difference


first row = a
second row = a+3
third row = a+3+3 = a+3(2)
fourth row = a+3+3+3 = a+3(3)
fifth row = a + 3+3+3+3 = a + 3(4)
and so on


nth row = a+3(n-1)
We have n = 52 rows in total
That 52nd row has 92 seats


nth row = a+3(n-1)
52nd row = a+3(52-1)
92 = a+3(52-1)
92 = a+3(51)
a+3(51) = 92
a+153 = 92
a = 92-153
a = -61


The arithmetic sequence is: {-61, -58, -55, ..., 86, 89, 92} which consists of 52 terms


As you can see, we run into a problem. There's no way to have a negative number of seats in the first row. Or any row really.
Your teacher made a typo somewhere.
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