if (b-c)2, (c-a)2, (a-b)2 are in AP, prove that 1/b-c, 1/c-a, 1/a-b are in AP|Arithmetic Progression

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if (b-c)^2, (c-a)^2, (a-b)^2 are in AP, prove that 1/b-c, 1/c-a, 1/a-b are in AP

Arithmetic Progression | Sequences and Series | Introduction | Infinite & Finite progression | Common difference | General term of a given AP | Sequence | Series

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Good morning sir, sir i have done this in a different method
If (b-c)^2, (c-a)^2, (a-b) is in ap
Therefore they square would also be in AP
i.e. (b-c), (c-a), (a-b) would also be in ap
If (b-c), (c-a), (a-b) is in ap then their reciprocal would also be in AP
Therefore 1/(b-c), 1/(c-a), 1/(a-b) is also in AP
Sir is this method correct?

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