The first row of an auditorium has 25 seats, and each successive row has 3 more seats. While finding the total seats in 20 rows, Meera writes: \(S_{20}=\frac{20}{2}[2(25)+20(3)]\). What is the correct correction in her solution?

Correct answer: \(20(3)\) should be replaced by \(19(3)\), because there are 19 gaps in 20 rows; the total is 1070 seats.

For the first 20 AP terms, \(S_n=\frac{n}{2}[2a+(n-1)d]\). Here \(n-1=19\), so \(S_{20}=10[50+57]=1070\). Twenty rows create only 19 increases. Exam tip: check the \((n-1)d\) term carefully.