Melaka Trial 2025 Add Maths Paper 2, Question 3Progressions

Solved on video in English · Form 4, Progressions · 6:19 video

The question

(a)

It is given that …, 145, x, y, 138, … is part of an arithmetic progression and the 16th term of the progression is 138.

(a) Find the first term and the common difference.

(b)

For the arithmetic progression with first term 173 and common difference −7/3 (from …, 145, x, y, 138, …):

(b) Find the smallest value of n such that the nth term is negative.

(c)

For the arithmetic progression with first term 173 and common difference −7/3 (from …, 145, x, y, 138, …):

Step 1

Since 138 is the 16th term and 145, x, y, 138 are consecutive terms, 145 is the 13th term (three terms before 138).

Why this step

Count the steps, not the terms. From 145 to 138 is three steps (through x and y), so 145 is the 13th term (16 − 3).

3 more steps, each with the reason for it, in the full solution

Final answer

a = 173, d = −7/3

Where students lose marks

Watch out: 145 is the 13th term, not the 14th. Take care with the sign: 138 − 145 = −7.

4 more steps, each with the reason for it, in the full solution

Final answer

n = 76

Where students lose marks

Watch out: n > 75.14 means n = 76 (the next whole number), not 75, because T₇₅ is still positive.

4 more steps, each with the reason for it, in the full solution

Final answer

1375

Where students lose marks

Watch out: subtract S₄₀, not S₄₁, to keep T₄₁ in the sum. Double-check 69(−7/3) = −161 and 39(−7/3) = −91.

Watch every step of this question solved on video (6:19), with the reason behind each step and a box that checks your own answer.

No card. Every past-year question on video, in English and BM.