The sum of the first (18) terms is (441), and the first term is (3). If the sequence is an AP, what is the common difference (d)?

Correct answer: \(\frac{43}{17}\)

For an AP, \(S_n=\frac{n}{2}[2a+(n-1)d]\). Thus, \(441=\frac{18}{2}[2(3)+17d]=9(6+17d)\). Hence \(49=6+17d\), so \(17d=43\) and \(d=\frac{43}{17}\). If \(d=3\), the sum would be \(513\), so it is not correct. Exam tip: first divide by \(\frac{n}{2}\) to simplify the equation.