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1998

JEE Main 1998 Mathematics question on Sequences and Series

If x>1,y>1,z>1x > 1, y > 1, z > 1 are in GP, then 11+ln⁡x,11+ln⁡y,11+ln⁡z\dfrac{1}{1 + \ln x}, \dfrac{1}{1 + \ln y}, \dfrac{1}{1 + \ln z} are in
  1. A.
    AP
  2. B.
    HP
  3. C.
    GP
  4. D.
    None of these
Show answer and solution

Correct option: B

Solution

Since x, y, z are in GP, we have ln x, ln y, ln z in AP. So ln x = a, ln y = a+d, ln z = a+2d for some common difference d.

The three terms become 1/(1+a), 1/(1+a+d), 1/(1+a+2d). Since (1+a), (1+a+d), (1+a+2d) are in AP, their reciprocals are in HP.

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