The first row of a theatre has (35) seats and each next row has (3) more seats. If the total number of seats is (840), how many rows are there?

Correct answer: (15)

The correct answer is C, 15 rows. The seat counts form an arithmetic progression with first term \\(a=35\\), common difference \\(d=3\\), and total \\(S_n=840\\). Use \\(S_n=\\frac{n}{2}[2a+(n-1)d]\\). Substitution gives \\(840=\\frac{n}{2}[70+3(n-1)]\\), so \\(840=\\frac{n}{2}(67+3n)\\). Multiplying by 2 gives \\(1680=n(67+3n)\\), or \\(3n^2+67n-1680=0\\). Testing the positive whole-number option, \\(n=15\\) gives \\(S_{15}=\\frac{15}{2}[70+14(3)]=\\frac{15}{2}(112)=840\\). Therefore C is correct. Option A gives \\(S_{13}=663\\), so it is too small. Option B gives \\(S_{14}=749\\), also too small. Option D gives \\(S_{16}=936\\), too large. The number of rows must be a positive whole number, not a fraction. Memory cue: for an AP seating problem, write first term, increase, total, then verify the chosen row count in the sum formula.