SPM 2017 Add Maths Paper 1, Question 7Progressions

Solved on video in English and Bahasa Melayu · 2 marks · Form 4, Progressions · 3:26 video

The question

It is given that the n-th term of a geometric progression is T_n = 3r^(n-1)/2, r != k. State (a) the value of k, (b) the first term of the progression.

[2 marks]

SPM 2017 Add Maths Paper 1, Question 7: the question as drawn in Eduly's video3:26

Step 1

For T_n = (3/2) r^(n-1) to be a valid geometric progression term, the common ratio r cannot take certain degenerate values; the mark scheme accepts k = 0 (ratio cannot be 0), or equivalently k = 1 or k = -1, as valid restricted values.

Why this step

If r = 0, all terms after the first are 0, so the ratio between terms is 0 ÷ 0, which is undefined. That's why k = 0.

1 more step, with the reason for it, in the full solution

Final answer

k = 0 (any one of k = 0, k = 1 or k = -1 is accepted); 3/2

Where students lose marks

Watch out: don't leave (a) blank (r = 0 breaks the ratio), and for (b) use n = 1, not n = 0.

Watch every step of this question solved on video (3:26), with the reason behind each step and a box that checks your own answer.

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