JEE Main 2023 · 31 January · morning · Numerical answer · +4 / −0

Let a₁, a₂, …, aₙ be in A.P. If a₅=2 a₇ and a₁₁=18, then
12(1/(√a₁₀+√a₁₁)+1/(√a₁₁+√a₁₂)+…+1/(√a₁₇+√a₁₈)) is equal to ____________.

Answer and solution

Answer 8

Solution

Maths sir · 4 steps

Rationalise each fraction, and the whole sum telescopes down to its first and last square roots.

  1. Find a and d. a₅ = 2a₇ gives a + 4d = 2a + 12d, so a = -8d. Then a₁₁ = a + 10d = 2d = 18, so d = 9 and a = -72.
  2. Terms you need. aₙ = -72 + 9(n-1) = 9n - 81. So a₁₀ = 9 and a₁₈ = 81, and every term from a₁₀ to a₁₈ is positive.

    1/(√aₖ+√aₖ₊₁) = (√aₖ₊₁-√aₖ)/(aₖ₊₁-aₖ) = (√aₖ₊₁-√aₖ)/d

  3. Telescope. Add this for k = 10 to 17. Every middle root cancels, and the sum is (√a₁₈-√a₁₀)/d = (9 - 3)/9 = 2/3.
  4. So. 12 × 2/3 = 8.

The common slip: 72. Forgetting to divide by d = 9 after rationalising leaves 12 × (9 - 3) = 72, not 8.

The idea to keep. A sum of 1/(√aₖ+√aₖ₊₁) over an A.P. always telescopes to (√last-√first)/d. Rationalise, and the common difference appears in the denominator.

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

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Asked about 250 times in JEE Main since 2020. This question is from 2023.