Let x₁, x₂, …, x₁₀₀ be in an arithmetic progression, with x₁=2 and their mean equal to 200 . If yᵢ=i(xᵢ-i), 1 ≤ i ≤ 100, then the mean of y₁, y₂, …, y₁₀₀ is :

Solution

Maths sir · 4 steps

The mean gives the last term and d; then split yᵢ into sums of i and i².

1. Find d. The mean of an AP is the average of its first and last terms, so (x₁ + x₁₀₀)/2 = 200 and x₁₀₀ = 398. Then 99d = 398 - 2 = 396, so d = 4.
2. Rewrite y. xᵢ = 2 + 4(i-1) = 4i - 2, so xᵢ - i = 3i - 2. That gives yᵢ = i(3i - 2) = 3i² - 2i.

   Σᵢ₌₁¹⁰⁰ yᵢ = 3Σᵢ₌₁¹⁰⁰ i² - 2Σᵢ₌₁¹⁰⁰ i = 3 × (100 × 101 × 201)/6 - 2 × (100 × 101)/2

3. Put in numbers. Σ i² = 338350 and Σ i = 5050. So the sum is 3 × 338350 - 2 × 5050 = 1015050 - 10100 = 1004950.
4. So. The mean of the yᵢ is 1004950/100 = 10049.5, which is option (b).

The idea to keep. For an AP, mean = (first + last)/2. When a new sequence is a polynomial in i, split it into Σ i and Σ i² and use the standard formulas.

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