Question 47 of 51intermediate🔧 ApplyNumerical3 marks

Which term of the sequence 2, 222\sqrt{2}, 4, … is 128?

Correct Answer

13th term

Exercise: Exercise Set 8.3 | Q: 6 | (Chapter: 21)
For More Understanding

Explanation

The given sequence 2, 22\sqrt{2}, 4, ... is a sequence where each term is obtained by multiplying the previous term by a constant multiplier (common ratio) of 2\sqrt{2}. Following the textbook's method of expressing terms using the common ratio, we can derive the explicit formula for the nth term. By equating this formula to 128 and expressing 128 as a power of 2\sqrt{2}, we can solve for the term number n.

Solution Steps

  1. Step 1: Identify the pattern and write the terms using the common ratio. The sequence can be rewritten as: t₁ = 2, t₂ = 2 × 2\sqrt{2}, t₃ = 2 × (2\sqrt{2})², t₄ = 2 × (2\sqrt{2})³, and so on.

  2. Step 2: Write the formula for the nth term. Following the pattern, the nth term is tₙ = 2 × (2\sqrt{2})ⁿ⁻¹.

  3. Step 3: Equate the nth term to 128 to find n. We have 2 × (2\sqrt{2})ⁿ⁻¹ = 128.

  4. Step 4: Simplify the equation. Dividing both sides by 2 gives (2\sqrt{2})ⁿ⁻¹ = 64.

  5. Step 5: Express 64 as a power of 2\sqrt{2}. Since 64 = 2⁶ and 2 = (2\sqrt{2})², we get 64 = (2\sqrt{2})¹².

  6. Step 6: Equate the exponents and solve for n. We get n - 1 = 12, which means n = 13. Thus, 128 is the 13th term.

PreviousNext