If log ⁡2,log ⁡2n−1 and 2n+3 are in AP, then n is equal to

If log 2,log 2n1 and 2n+3 are in AP, then n is equal to

  1. A

    52

  2. B

    log2 5

  3. C

    log3 5

  4. D

    32

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    Solution:

    Since, 1og2 , log(2n -1) and log (2n+ 3) are in AP. 

    2log2n1=log2+log2n+32n12=22n+32n52n+1=0

    As 2n cannot be negative hence, 

    2n5=0 2n=5n=log25

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