All of the seats in a lecture hall are arranged in 15 rows. After the front row, each row has one more seat than the row in front of it. How many seats are in the 8th row from the front? (1) The lecture hall has a total of 285 seats. (2) The last row in the back has 26 seats.
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This is a nice arithmetic sequence setup. Took me a second to realize we need both the first term and the common difference to pin down the 8th row.
Same here. I almost picked A before realizing one equation can still leave two unknowns.
Yeah, but each statement actually gives you a different equation, so it's worth checking them separately.
For statement 2, doesn't the last row being 26 basically give us the common difference since there are 15 rows? First row = 26 - 14 = 12, then just count up to the 8th.
That's what I did too. Way faster than messing with the sum formula.
Wait, for statement 1 I set up the sum formula but got stuck. Can someone explain how the total 285 lets you solve for the first row?
Use sum = n/2 * (first + last) and last = first + 14. Two equations, two variables.
Or 15/2 * (2a + 14) = 285. Solve for a.
Got D pretty quickly. 505-555 feels right, not too tricky once you spot the sequence.
What it tests
Your ability to spot and extend arithmetic and geometric sequences and work with recursive definitions.
Common trap
Off-by-one errors in indexing (a₀ vs a₁) or confusing an arithmetic difference with a geometric ratio.