Find the nth term of the AP: 11, 8, 5, 2 … Write the recursive rule for this AP.
Answer: tn = 14 − 3n. Recursive rule: t1 = 11, tn = tn−1 − 3 for n ≥ 2.
Step-by-step solution
Idea: Find a and d. The explicit rule is a + (n − 1)d; the recursive rule is ‘first term a, then add d each time’.
- a = 11, d = 8 − 11 = −3.½ mark
- tn = 11 + (n − 1)(−3) = 11 − 3n + 3 = 14 − 3n.½ mark
- Recursive rule: t1 = 11, tn = tn−1 − 3 for n ≥ 2.1 mark
tₙ = 14 − 3n; recursive rule t₁ = 11, tₙ = tₙ₋₁ − 3 for n ≥ 2.
Check: 14 − 3 × 4 = 2 ✓ (4th term); recursion 11 → 8 → 5 → 2 ✓.
Answer to write in the exam
a = 11, d = 8 − 11 = −3
tn = 11 + (n − 1)(−3)
∴ tn = 14 − 3n
Recursive rule: t1 = 11, tn = tn−1 − 3 for n ≥ 2
Common mistakes that cost marks
- Expanding (n − 1)(−3) as −3n − 3. It is −3n + 3, giving 14 − 3n, not 8 − 3n.
- Writing the recursive rule without the first term, or as tn = tn−1 + 3.
- Writing 3n − 14 (the signs reversed); check with n = 1: it must give 11.
How this can come in the exam
MCQ (1 mark)
The nth term of the AP 20, 15, 10, 5, … is
- 25 − 5n
- 20 − 5n
- 5n + 15
- 15 − 5n
Show answer
(A) 25 − 5n
20 + (n − 1)(−5) = 25 − 5n.
Try one yourself
Find the nth term of the AP 9, 2, −5, −12, … and write its recursive rule.
Show answer
d = −7: tn = 16 − 7n; t1 = 9, tn = tn−1 − 7 for n ≥ 2.
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