Find the first term and the common difference.

Find the smallest value of n such that the sum of the first n terms is greater than 600.

\({S_n} = \frac{n}{2}[2a + (n - 1)d]\)

\(212 = \frac{{16}}{2}(2a + 15d)\,\,\,\,\,( = 16a + 120d)\)     A1

\({n^{th}}{\text{ term is }}a + (n - 1)d\)

\(8 = a + 4d\)     A1

solving simultaneously:     (M1) 

\(d = 1.5,{\text{ }}a = 2\)     A1

[4 marks]

\(\frac{n}{2}[4 + 1.5(n - 1)] > 600\)     (M1)

\( \Rightarrow 3{n^2} + 5n - 2400 > 0\)     (A1)

\( \Rightarrow n > 27.4...,{\text{ }}(n < - 29.1...)\) 

Note: Do not penalize improper use of inequalities. 

\( \Rightarrow n = 28\)     A1

[3 marks]

This proved to be a good start to the paper for most candidates. The vast majority made a meaningful attempt at this question with many gaining the correct answers. Candidates who lost marks usually did so because of mistakes in the working. In part (b) the most efficient way of gaining the answer was to use the calculator once the initial inequality was set up. A small number of candidates spent valuable time unnecessarily manipulating the algebra before moving to the calculator.

This proved to be a good start to the paper for most candidates. The vast majority made a meaningful attempt at this question with many gaining the correct answers. Candidates who lost marks usually did so because of mistakes in the working. In part (b) the most efficient way of gaining the answer was to use the calculator once the initial inequality was set up. A small number of candidates spent valuable time unnecessarily manipulating the algebra before moving to the calculator.