Let a₁, a₂, …, a₂₀₂₄ be an Arithmetic Progression such that a₁+(a₅+a₁₀+a₁₅+…+a₂₀₂₀)+a₂₀₂₄=2233. Then a₁+a₂+a₃+…+a₂₀₂₄ is equal to _________.

Solution

Maths sir · 4 steps

Pair terms whose positions add to 2025, because every such pair equals a₁ + a₂₀₂₄.

1. Equal-sum pairs. In an AP, aₚ + a_q = a₁ + a₂₀₂₄ whenever p + q = 1 + 2024 = 2025. Inside the bracket, 5 + 2020 = 2025, 10 + 2015 = 2025, and so on.
2. Count the terms. The bracket has a₅, a₁₀, …, a₂₀₂₀, which is 2020/5 = 404 terms. They make 202 pairs, and each pair equals a₁ + a₂₀₂₄.

   (a₁ + a₂₀₂₄) + 202 (a₁ + a₂₀₂₄) = 203 (a₁ + a₂₀₂₄) = 2233

3. Find the end sum. a₁ + a₂₀₂₄ = 2233/203 = 11.
4. So. S₂₀₂₄ = 2024/2 (a₁ + a₂₀₂₄) = 1012 × 11 = 11132.

The common slip: 22264. Forgetting the 1/2 in Sₙ = n/2(a₁ + aₙ) gives 2024 × 11 = 22264, not 11132.

The idea to keep. When an AP sum hides scattered terms, look for positions that add to the same total. Each such pair equals first term plus last term, and that one number usually gives the whole sum.

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