If { aᵢ} ᵢ₌₁ⁿ, where n is an even integer, is an arithmetic progression with common difference 1, and Σᵢ₌₁ⁿ aᵢ = 192 , Σᵢ₌₁^(n/2) a₂ᵢ = 120 , then n is equal to :

Solution

Maths sir · 3 steps

Split the terms into odd and even places, then pair each odd term with the even term right after it.

1. Odd-place sum. The even-place terms add to 120 and all terms add to 192. So the odd-place terms a₁, a₃, …, aₙ₋₁ add to 192 - 120 = 72.
2. Pair them up. Since n is even, the terms form n/2 pairs (a₁, a₂), (a₃, a₄), …. In each pair the even term is bigger by the common difference, which is 1.

   Σᵢ₌₁^(n/2) a₂ᵢ - Σᵢ₌₁^(n/2) a₂ᵢ₋₁ = n/2 × 1

3. So. 120 - 72 = 48 = n/2, which gives n = 96, option (b).

The idea to keep. When an AP is split into odd and even places, subtract the two sums. Each pair differs by d, so the gap is (number of pairs) × d. You do not need the first term at all.

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