If a1,a2,a3,…,a4001are term of an AP such  that 1a1a2+1a2a3+…+1a4000a4001=10a2 and  a2+a4000=50, then a1−a4001is equal to

If a1,a2,a3,,a4001are term of an AP such  that 1a1a2+1a2a3++1a4000a4001=10a2 and  a2+a4000=50, then a1a4001is equal to

  1. A

    20

  2. B

    30

  3. C

    40

  4. D

    None of these 

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    Solution:

    Now,1a1a2+1a2a3++1a4000a4001

    =1da2a1a1a2+a3a2a2a3++a4001a4000a4000a4001=1d1a11a2+1a21a3++1a40001a4001

    =1d1a11a4001=4000a1a4001=10       (given)  a1a4001=400                              ...(i)            a1+a4001=a2+a4000=50                .....(ii) a1a40012=a1+a400124a1a4001=(50)21600 a1a4001=30

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