2015 PAPER 1 – QUESTION 6.

This exam question examines students’ knowledge of the Tn and Sn formulas. Then general term of an arithmetic sequence

∈ is T =15–2 n, where n N.

(i) Write down the first three terms of the sequence.

The first three terms are also known as T1, T2 and T3. Substitute n =1, n =2and n = 3 into the Tn formula.

T =15 – 2(1)=15 – 2=13

T1=15 – 2(2)=15 – 4=11

T2=15 – 2(3)=15 – 6=9

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(ii) Find the first negative term of the sequence.

The sequence above is an arithmetic sequence since the difference between each term is the same. This sequence is decreasing by 2 with each term. Continue the pattern until you reach the first negative number.

13, 11, 9, 7, 5, 3, 1, – 1. Therefore, the eighth term contains the first negative term of the sequence, i.e. T8 = – 1.

(i) Find Sn = T1 + T2 + … + Tn , the sum of the first n terms of the series, in terms of n.

First of all, highlight ‘in terms of n’. This tells the student that their answer will have n in it. This question wants students to find a general formula for the sum of the first n terms. Start these questions by filling in the values of a and d. The first term is a = 13 and the difference is d = – 2. Sub these values into the Sn formula: (ii) Find the value of n for which the sum of the first n terms of the series is 0.

The question wants the student to use the formula Sn = 0. The sum of the first n terms cannot be zero. Therefore, the answer is 14.

Finally, what if there was a part where they wanted the student to find the sum of the first

100 terms? All you have to do is substitute n =

100 into the Sn formula.