Prove that a,b,c are in A.P. iff 1/bc,1/ca,1/ab are also in A.P.

Consider 1/bc, 1/ca, 1/ab are in AP,

So, 1/ca = (1/bc + 1/ab)/2

or, 1/ca = (a+c)/2abc

or, 1/ca = (a+c)/2abc

or, 1 = (a+c)/2b

or, b = (a+c)/2

Thus, 'b' is A.M. of  'a' and 'c',

∴ a, b, c are in A.P. if 1/bc, 1/ca, 1/ab are in A.P.


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when 1/bc , 1/ca ,1/ab are in AP 

  1/ca - 1/bc = 1/ab - 1/ca

bc -ca/abc2   = ca - ab / a2 bc

c(b-a)/ c = a(c-b)/ a 

 then it becomes

b-a = c-b

 i.e 2b = a+c

therefore a, b,c are in AP

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If 1/bc, 1/ca, 1/ab in A.P. So, abc/bc, abc/ca, abc/ab is also in A.P. (since, we are multiplying the above one with abc). Therefore, a, b, c is in A.P. [Proved]
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if a b c are in a.p (a/bc) (b+c) (b/ca) (c+a) (c/ab) (a+b) are in.
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