(ii) tn = 2n – 5 (1 mark)


(ii) tn = 2n – 5 (1 mark)

Sol. tn = 2n – 5
t1 = 2 (1) – 5 = 2 – 5 = – 3
t2 = 2 (2) – 5 = 4 – 5 = – 1
t3 = 2 (3) – 5 = 6 – 5 = 1
t4 = 2 (4) – 5 = 8 – 5 = 3
t5 = 2 (5) – 5 = 10 – 5 = 5

The first five terms of the sequence are – 3, – 1, 1, 3 and 5.

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