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Question
If (a − b), (b − c), (c − a) are in G.P., then prove that (a + b + c)2 = 3 (ab + bc + ca)
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Solution
\[\left( a - b \right), \left( b - c \right) \text { and }\left( c - a \right) \text { are in G . P} . \]
\[ \therefore \left( b - c \right)^2 = \left( a - b \right)\left( c - a \right)\]
\[ \Rightarrow b^2 - 2bc + c^2 = ac - bc + ab - a^2 \]
\[ \Rightarrow a^2 + b^2 + c^2 = ab + bc + ca . . . . . . . (i)\]
\[\text{ Now, LHS } = \left( a + b + c \right)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\]
\[ = ab + bc + ca + 2ab + 2bc + 2ca \left[\text { Using }(i) \right]\]
\[ = 3ab + 3bc + 3ca\]
\[ = 3\left( ab + bc + ca \right)\]
\[ = \text { RHS }\]
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