Suppose 12x,|x+1|,|x−1| are in A.P., then sum to 10 terms of the A.P. is

Suppose 12x,|x+1|,|x1| are in A.P., then sum to 10 terms of the A.P. is

  1. A

    54

  2. B

    36

  3. C

    28

  4. D

    none of these

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    Solution:

    As 12x,|x+1|,|x1| are in A.P.

    2|x+1|=12x+|x1|                         (1)

    If x<1then (1) give us 

    2(x+1)=12x(x1)

    x=2

    In this case common difference is 2 and sum to 10 terms is 

    102[2(1)+(9)(2)]=85

    If  1x<1 

    2(x+1)=12x(x1)x=2/5

    In this case common difference is 4/5, and sum to 10 terms is

    102215+(9)45=34

    If x1,then (1) becomes 

    2x+2=12x+x1

    12x=3

    Not possible as x1

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