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Question
If a, b, c are in G.P., prove that:
\[\frac{(a + b + c )^2}{a^2 + b^2 + c^2} = \frac{a + b + c}{a - b + c}\]
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Solution
a, b and c are in G.P.
\[\therefore b^2 = ac\] .......(1)
\[\text { LHS }= \frac{\left( a + b + c \right)^2}{a^2 + b^2 + c^2}\]
\[ = \frac{\left( a + b + c \right)^2}{a^2 - b^2 + c^2 + 2 b^2}\]
\[ = \frac{\left( a + b + c \right)^2}{a^2 - b^2 + c^2 + 2ac} \left[ \text { Using } (1) \right]\]
\[ = \frac{\left( a + b + c \right)^2}{\left( a + b + c \right)\left( a - b + c \right)} \left[ \because \left( a + b + c \right)\left( a - b + c \right) = a^2 - b^2 + c^2 + 2ac \right]\]
\[ = \frac{\left( a + b + c \right)}{\left( a - b + c \right)} =\text { RHS }\]
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