Posted by **Anonymous** on Wednesday, September 21, 2011 at 11:47am.

Chairs in an auditorium are arranged in rows in such a way that the first two rows each have the same number of chairs. The third and fourth rows each have three more chairs than the first and second row; the fifth and sixth rows each have three more chairs than the third and fourth row, etc. The sequence of number of chairs for every second row forms an arithmetic sequence. The first two rows each have 27 chairs, and the last two rows each have 114 chairs.

How many rows of chairs are there?

How can this be? Because the first two rows have the same amount of chairs, and no whole number added by itself equals 27. I'm Confused.

- Calculus -
**Steve**, Wednesday, September 21, 2011 at 1:21pm
The first 2 rows *each* have 27 chairs.

What we have here is the fact that each pair of rows of chairs has 3 more per row more than the previous pair.

114-27 = 87 = 3*29.

So, there are 30 pairs of rows. The first pair have 27 chairs each, and the 30th pair have 114 chairs each.

Sounds like 60 rows of chairs.

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