Can you find a recursive rule for the formula t_n = 3 × 10ⁿ⁻¹ that generates the geometric progression 3, 30, 300, 3000, … ?

Step-by-step solution

Idea: In a GP each term is the previous term times the common ratio. Here r = 303 = 10 and the first term is 3.

1. t₁ = 3 × 10⁰ = 3. Ratio: t_nt_n−1 = 3 × 10ⁿ⁻¹3 × 10ⁿ⁻² = 10.1 mark
2. So the recursive rule is t₁ = 3, t_n = 10 × t_n−1 for n ≥ 2.1 mark

Check: 3 → 30 → 300 → 3000 ✓.

Answer to write in the exam

Common mistakes that cost marks

• Writing t_n = t_n−1 + 10 (an AP rule). In a GP you multiply by the ratio.
• Leaving out the first term 3; the rule then does not fix the sequence.
• Writing t_n = 3t_n−1. The ratio is 10; 3 is the first term.

How this can come in the exam

Try one yourself

Write a recursive rule for t_n = 5 × 4ⁿ⁻¹ and its first four terms.

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