A sequence is given by the recursive rule t1 = 2, tn+1 = 3tn − 2 for n ≥ 1. Which term of the sequence is 730?
Answer: Terms: 2, 4, 10, 28, 82, 244, 730, … so 730 is the 7th term. (In fact tn = 3n−1 + 1, and 730 = 36 + 1.)
Step-by-step solution
Idea: Generate the terms with the rule until 730 appears. A neat check: each term is 1 more than a power of 3.
- t2 = 3 × 2 − 2 = 4; t3 = 3 × 4 − 2 = 10; t4 = 3 × 10 − 2 = 28.1 mark
- t5 = 3 × 28 − 2 = 82; t6 = 3 × 82 − 2 = 244; t7 = 3 × 244 − 2 = 730.1 mark
- So 730 is the 7th term. Pattern: subtracting 1 from each term gives 1, 3, 9, 27, 81, 243, 729 = 30, …, 36, so tn = 3n−1 + 1 and 730 = 36 + 1 = t7.1 mark
730 is the 7th term.
Check: 3⁶ + 1 = 729 + 1 = 730 ✓.
Answer to write in the exam
t2 = 3 × 2 − 2 = 4
t3 = 3 × 4 − 2 = 10
t4 = 3 × 10 − 2 = 28
t5 = 3 × 28 − 2 = 82
t6 = 3 × 82 − 2 = 244
t7 = 3 × 244 − 2 = 730
∴ 730 is the 7th term.
Common mistakes that cost marks
- Computing 3tn − 2 as 3(tn − 2): 3 × (2 − 2) = 0. Multiply first, then subtract 2.
- Counting the steps (6) instead of the term number (7): the first term is 2.
- Treating the sequence as a GP with ratio 3 (2, 6, 18, …). The −2 changes every term.
How this can come in the exam
MCQ (1 mark)
If t1 = 1 and tn+1 = 3tn + 1 for n ≥ 1, then t5 is
- 40
- 121
- 364
- 81
Show answer
(B) 121
1, 4, 13, 40, 121.
Try one yourself
t1 = 1 and tn+1 = 2tn + 1 for n ≥ 1. Which term is 255?
Show answer
1, 3, 7, 15, 31, 63, 127, 255: the 8th term (each term is 2n − 1).
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