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1999

JEE Main 1999 Mathematics question on Sequences and Series

Let a1,a2,…,a10a_1, a_2, \ldots, a_{10} be in AP and h1,h2h_1, h_2, equal to …,h10\ldots, h_{10} be in HP. If a1=h1=2a_1 = h_1 = 2 and a10=h10=3a_{10} = h_{10} = 3, then a4h7a_4 h_7 is
  1. A.
    2
  2. B.
    3
  3. C.
    5
  4. D.
    6
Show answer and solution

Correct option: D

Solution

For the AP: a₁=2, a₁₀=3, so common difference d = (3-2)/9 = 1/9. Thus a₄ = 2 + 3d = 2 + 3/9 = 7/3.

For the HP: h₁=2, h₁₀=3, so 1/h₁, 1/h₂,…,1/h₁₀ are in AP with first term 1/2 and last term 1/3. Common difference D = (1/3 - 1/2)/9 = (-1/6)/9 = -1/54. Thus 1/h₇ = 1/2 + 6×(-1/54) = 1/2 - 6/54 = 1/2 - 1/9 = 7/18, giving h₇ = 18/7.

Therefore a₄·h₇ = (7/3)×(18/7) = 18/3 = 6.

[1999]

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