Which term of the sequence 2, 2√2, 4, … is 128?

Step-by-step solution

Idea: 2√22 = √2 and 42√2 = √2, so the ratio is √2. Write 64 as a power of √2 using (√2)² = 2.

1. r = 2√22 = √2 and 42√2 = 2√2 = √2. So it is a GP with a = 2, r = √2.1 mark
2. t_n = 2 × (√2)ⁿ⁻¹ = 128 ⇒ (√2)ⁿ⁻¹ = 64.½ mark
3. 64 = 2⁶ = ((√2)²)⁶ = (√2)¹², so n − 1 = 12.1 mark
4. n = 13. 128 is the 13th term.½ mark

Check: Every two steps multiply by (√2)² = 2: odd-numbered terms are 2, 4, 8, 16, 32, 64, 128 for n = 1, 3, 5, 7, 9, 11, 13 ✓.

Answer to write in the exam

Common mistakes that cost marks

• Writing 64 = 2⁶ and concluding n − 1 = 6. The base is √2, not 2, so the exponent doubles to 12.
• Taking r = 2 because the third term 4 is double the first.
• Forgetting to add 1 at the end and answering 12.

How this can come in the exam

Try one yourself

Which term of the GP 3, 3√3, 9, … is 243?

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