This question was also reported to being very popular among the candidates. However majority of the candidates were reported to have performed poorly in the question.
In part (a), they could not obtain the three equations involved. This the report attributed to their inability to manipulate the algebraic expressions involved. Candidates were expected to recall the nth term of an A.P. as Tn = a + (n - l)d, where a = first term, n= the term sought, d = common difference and Tn = value of the term sought. T2 = a + d. This implied that x - 1 = a + d. T4 = a + 3d x + 1 = a + 3d. T4 - T2 = 2d = [(x + 1) - (x - 1)] = 2. Hence 2d = 2 d = 1. T6 = 7(given). i.e. a +5d = 7. Simplifying gave a = 2. Substituting 2 for a and 1 for d in T2, 3 = x - 1x = 4.
In part (b), candidates' performance was reported to be poorer than it was in part (a). Majority of the candidates did not get the required answer mainly because they did not read the question carefully to understand what was required. Majority of them took the sphere to be filled when 6 litres of water was poured into it instead of one-quarter filled as stated in the question, hence they lost some marks. Others did not convert the unit from litres to cm³ before solving. Since the sphere was one-quarter full, then ¼ x 4/3 X22/7X r3 = 6000cm³  
Solving this equation gave r = 17.9 cm. Hence diameter = 2 x r = 35.8 cm correct to 3 significant figures.
The Chief Examiner reported that responses were not received from candidates for questions 14 and 15 as these questions were meant for candidates in Ghana, Sierra Leone and The Gambia.