If 1b−c,1c−a,1a−bbe consecutive terms of an AP, then (b−c)2,(c−a)2,(a−b)2 will be in 

If 1bc,1ca,1abbe consecutive terms of an AP, then (bc)2,(ca)2,(ab)2 will be in 

  1. A

    GP

  2. B

    AP

  3. C

    HP

  4. D

    none of these

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    Solution:

    Now,  we assume (bc)2,(ca)2,(ab)2 are in AP, then we have

    (ca)2(bc)2=(ab)2(ca)2 (ba)(2cab)=(cb)(2abc)    ......(i)

    Also,if  1bc,1ca,1ab are in,AP, then

    1ca1bc=1ab1cab+a2c(ca)(bc)=c+b2a(ab)(ca) (ab)(b+a2c)=(bc)(c+b2a) (ba)(2cab)=(cb)(2abc)

    which is equal to Eq.(i), so our hypothesis is true.

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