Write down the value of the common difference.

Find the first term.

Find the sum of the first 50 terms of the sequence.

\(d =  - 1.5\)     A1     N1

[1 mark]

METHOD 1

valid approach     (M1)

eg\(\;\;\;{u_{10}} = {u_1} + 9d,{\text{ }}8 = {u_1} - 9( - 1.5)\)

correct working     (A1)

eg\(\;\;\;8 = {u_1} + 9d,{\text{ }}6.5 = {u_1} + 10d,{\text{ }}{u_1} = 8 - 9( - 1.5)\)

\({u_1} = 21.5\)     A1     N2

METHOD 2

attempt to list 3 or more terms in either direction     (M1)

eg\(\;\;\;9.5,{\text{ }}11,{\text{ }}12.5,{\text{ }} \ldots ;{\text{ }}5,{\text{ }}3.5,{\text{ }}2,{\text{ }} \ldots {\text{ }} \ldots \)

correct list of 4 or more terms in correct direction     (A1)

eg\(\;\;\;9.5,{\text{ }}11,{\text{ }}12.5,{\text{ }}14\)

\({u_1} = 21.5\)     A1     N2

[3 marks]

correct expression     (A1)

eg\(\;\;\;\frac{{50}}{2}\left( {2(21.5) + 49( - 1.5)} \right),{\text{ }}\frac{{50}}{2}(21.5 - 52),{\text{ }}\sum\limits_{k = 1}^{50} {21.5 + (k - 1)( - 1.5)} \)

\({\text{sum}} =  - 762.5\;\;\;{\text{(exact)}}\)     A1     N2

[2 marks]

Total [6 marks]

In general, candidates showed confidence in this area of the syllabus. Appropriate formulae were chosen for parts (b) and (c) and many candidates were able to achieve full marks. However, many candidates found the common difference to be \( + 1.5\) in part (a) by subtracting \({u_{10}} - {u_{11}}\) or believing that the common difference should always be positive.

In general, candidates showed confidence in this area of the syllabus. Appropriate formulae were chosen for parts (b) and (c) and many candidates were able to achieve full marks. However, many candidates found the common difference to be \( + 1.5\) in part (a) by subtracting \({u_{10}} - {u_{11}}\) or believing that the common difference should always be positive.

In general, candidates showed confidence in this area of the syllabus. Appropriate formulae were chosen for parts (b) and (c) and many candidates were able to achieve full marks. However, many candidates found the common difference to be \( + 1.5\) in part (a) by subtracting \({u_{10}} - {u_{11}}\) or believing that the common difference should always be positive.