If \[a,\,b,\,c\] are in A.P., then \[(a+2b-c)\]\[(2b+c-a)\]\[(c+a-b)\] equals [Pb. CET 1999]

If \[a,\,b,\,c\] are in A.P., then \[(a+2b-c)\]\[(2b+c-a)\]\[(c+a-b)\]

1.	If \[a,\,b,\,c\] are in A.P., then \[(a+2b-c)\]\[(2b+c-a)\]\[(c+a-b)\] equals [Pb. CET 1999]
A.	\[\frac{1}{2}abc\]
B.	abc
C.	2 abc
D.	4 abc
Answer» E.

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If \[a,\,b,\,c\] are in A.P., then \[(a+2b-c)\]\[(2b+c-a)\]\[(c+a-b)\] equals [Pb. CET 1999]

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