JEE Main 2023 · 6 April · morning · Single correct · +4 / −1
Let a₁, a₂, a₃, …, aₙ be n positive consecutive terms of an arithmetic progression. If d > 0 is its common difference, then
lim n → ∞ √(d/n)(1/(√a₁+√a₂)+1/(√a₂+√a₃)+… … …+1/(√aₙ₋₁+√aₙ)) is
- (a) 1/√d
- (b) 1
- (c) 0
- (d) √d
Answer and solution
Answer (b) 1
Solution
Maths sir · 4 steps
Rationalise each term and the whole sum telescopes down to just the first and last terms.
- Rationalise one term. Multiply the top and bottom of 1/(√aₖ+√aₖ₊₁) by √aₖ₊₁-√aₖ. The denominator becomes aₖ₊₁-aₖ = d, so each term equals (√aₖ₊₁-√aₖ)/d.
- Telescope the sum. Adding these from k=1 to n-1, every middle square root cancels. Only (√aₙ-√a₁)/d is left.
√(d/n)·(√aₙ-√a₁)/d = √(a₁+(n-1)d)/√nd - √a₁/√nd
- Take the limit. The first part is √((a₁+(n-1)d)/nd) = √((a₁-d)/nd+1) → √1 = 1. The second part has a fixed numerator over √nd → ∞, so it goes to 0.
- So. The limit is 1 - 0 = 1, which is option (b).
The idea to keep. When terms look like 1/(√aₖ+√aₖ₊₁) in an AP, rationalise. The difference of consecutive terms is d, and the sum collapses to the first and last terms.
Answer from the NTA final answer key. Solution by PrepWiser.