Let x₁, x₂, x₃, …, x₂₀ be in geometric progression with x₁=3 and the common ratio 1/2. A new data is constructed replacing each xᵢ by (xᵢ-i)². If x̄ is the mean of new data, then the greatest integer less than or equal to x̄ is ____________.

Solution

Maths sir · 4 steps

Expand the square into three sums. Two are standard GPs, and Σ i² does almost all the work.

1. Split the sum. Here xᵢ = 3(1/2)ⁱ⁻¹. Expand each new value as (xᵢ - i)² = xᵢ² - 2i xᵢ + i² and add the three parts separately.

   Σᵢ₌₁²⁰(xᵢ-i)² = Σ xᵢ² - 2Σ i xᵢ + Σ i²

2. Three sums. xᵢ² is a GP with first term 9 and ratio 1/4, so Σ xᵢ² = 12(1 - 4⁻²⁰). The arithmetico-geometric sum is Σ i(1/2)ⁱ⁻¹ = 4 - 44/2²⁰, so Σ i xᵢ = 12 - 132/2²⁰. Finally Σ i² = (20 × 21 × 41)/6 = 2870.
3. Add them up. Total = 12 - 24 + 2870 + 264/2²⁰ - 12/4²⁰ = 2858 + 264/2²⁰ - 12/4²⁰. The extra fractions add up to only about 0.00025.
4. So. x̄ = 2858.00025/20 ≈ 142.90001, so the greatest integer less than or equal to x̄ is 142.

The common slip: 144. Dropping the cross term -2Σ i xᵢ gives (12 + 2870)/20 = 144.1, so the answer comes out as 144 instead of 142.

The idea to keep. For the mean of (aᵢ - bᵢ)², expand the square and find each sum on its own. Tiny tail terms like 1/2²⁰ almost never change a greatest-integer answer.

Answer from the NTA final answer key. Solution by PrepWiser.