Observation

The Chief Examiner reported that few of the Candidates could not translate the word problem into equation as they used the nth term and not the sum of the first ten terms. As a result of this, the simultaneous equation they obtained and the solutions were wrong and thereby losing the marks allocated.
In part (a), it was expected that they substitute into the formula for sum of an Arithmetic Progression to have 130 = 10/2 [2a+(10-1)d] and simplifying will yield 2a + 9d = 26 -------(1).
Also, U5 = a + (5 – 1)d = 3a and simplifying will yield a = 2d --------------------------------(2) Substituting for a in equation (1) will yield 2(2d) + 9d = 26 and solving for d, d = 2. In part (b), Substituting for d in equation (2) to get the first term:a = 2(2) = 4 Then in part (c), the number of terms required can be obtained by solving 28 = 4 + (n – 1)2 which resulted to n = 13.