Arithmetic progressions · 2 marks

Find recursive rules for the APs in the previous exercises.

Answer: 2, 5, 8, …: t₁ = 2, t_n = t_n−1 + 3. −5, −1, 3, …: t₁ = −5, t_n = t_n−1 + 4. 12, 52, …: t₁ = 12, t_n = t_n−1 + 2. 1.5, 3.5, …: t₁ = 1.5, t_n = t_n−1 + 2. (All for n ≥ 2.)

Step-by-step solution

Given: The four APs from the previous exercises: 2, 5, 8, 11, …; −5, −1, 3, 7, …; 12, 52, 92, 132, …; 1.5, 3.5, 5.5, 7.5, …

Idea: The recursive rule of an AP is t₁ = a, t_n = t_n−1 + d for n ≥ 2. So for each AP we only need its first term and its common difference.

2, 5, 8, 11, …: a = 2, d = 3 ⇒ t₁ = 2, t_n = t_n−1 + 3 for n ≥ 2.½ mark
−5, −1, 3, 7, …: a = −5, d = 4 ⇒ t₁ = −5, t_n = t_n−1 + 4 for n ≥ 2.½ mark
12, 52, 92, …: a = 12, d = 2 ⇒ t₁ = 12, t_n = t_n−1 + 2 for n ≥ 2.½ mark
1.5, 3.5, 5.5, …: a = 1.5, d = 2 ⇒ t₁ = 1.5, t_n = t_n−1 + 2 for n ≥ 2.½ mark

t₁ = 2, tₙ = tₙ₋₁ + 3; t₁ = −5, tₙ = tₙ₋₁ + 4; t₁ = 1/2, tₙ = tₙ₋₁ + 2; t₁ = 1.5, tₙ = tₙ₋₁ + 2 (each for n ≥ 2).

Answer to write in the exam

Recursive rule of an AP: t₁ = a, t_n = t_n−1 + d (n ≥ 2)

2, 5, 8, 11, …: t₁ = 2, t_n = t_n−1 + 3

−5, −1, 3, 7, …: t₁ = −5, t_n = t_n−1 + 4

12, 52, 92, …: t₁ = 12, t_n = t_n−1 + 2

1.5, 3.5, 5.5, …: t₁ = 1.5, t_n = t_n−1 + 2

Common mistakes that cost marks

Leaving out the first term. ‘Add 3 each time’ alone does not say where to start; the rule needs both t₁ and the step.
Writing the rule as t_n = t_n−1 + d for n ≥ 1. There is no t₀, so the rule starts at n = 2.
Writing the step as the ratio of terms (52 ÷ 12 = 5) instead of the difference: d = 52 − 12 = 2.

How this can come in the exam

MCQ (1 mark)

A recursive rule for the AP 15, 11, 7, 3, … is

t₁ = 15, t_n = t_n−1 + 4
t₁ = 15, t_n = t_n−1 − 4
t₁ = 3, t_n = t_n−1 + 4
t₁ = 15, t_n = 4t_n−1

Show answer

(B) t₁ = 15, t_n = t_n−1 − 4
First term 15 and common difference −4.

Try one yourself

Write a recursive rule for the AP 0.2, 0.9, 1.6, 2.3, …