Let Sₙ=1/2+1/6+1/12+1/20+… upto n terms. If the sum of the first six terms of an A.P. with first term -p and common difference p is √(2026 S₂₀₂₅), then the absolute difference betwen 20ᵗʰ and 15ᵗʰ terms of the A.P. is

Solution

Maths sir · 4 steps

Split each term as 1/k-1/(k+1), let the series collapse, then the A.P. gives p.

1. Spot the pattern. The denominators are 2, 6, 12, 20, which are 1×2, 2×3, 3×4, 4×5. So the k-th term is 1/k(k+1) = 1/k-1/(k+1).
2. Telescope. When you add these, every middle fraction cancels with its neighbour. Only the first and last pieces remain, so Sₙ = 1-1/(n+1) = n/(n+1).

   √(2026 S₂₀₂₅) = √(2026×2025/2026) = √2025 = 45

3. Find p. The A.P. has a=-p and d=p, so the sum of its first six terms is 6/2[2(-p)+5p] = 9p. Setting 9p = 45 gives p = 5.
4. So. T₂₀-T₁₅ = (20-15)d = 5p = 25. The absolute difference is 25, option (d).

The idea to keep. When the denominator is a product of consecutive numbers, split the term into a difference of two fractions. The middle terms cancel and only the two ends survive.

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