If \(S_n\) is the sum of the first n terms of an arithmetic progression, which relation gives the nth term?

Correct answer: \(a_n=S_n-S_{n-1}\)

\(S_n=a_1+a_2+\cdots+a_{n-1}+a_n\), whereas \(S_{n-1}=a_1+a_2+\cdots+a_{n-1}\). Thus, subtracting \(S_{n-1}\) from \(S_n\) leaves only the nth term: \(a_n=S_n-S_{n-1}\). \(S_n/n\) is the average of the first n terms, not generally the nth term. Exam tip: obtain a term from consecutive sums by subtracting the previous sum from the larger sum.