Question

If a, b, c are in G.P., prove that:

\[\frac{(a + b + c )^2}{a^2 + b^2 + c^2} = \frac{a + b + c}{a - b + c}\]

Answer 1

a, b and c are in G.P.

\[\therefore b^2 = ac\]   .......(1)

\[\text {  LHS  }= \frac{\left( a + b + c \right)^2}{a^2 + b^2 + c^2}\]

\[ = \frac{\left( a + b + c \right)^2}{a^2 - b^2 + c^2 + 2 b^2}\]

\[ = \frac{\left( a + b + c \right)^2}{a^2 - b^2 + c^2 + 2ac} \left[ \text { Using } (1) \right]\]

\[ = \frac{\left( a + b + c \right)^2}{\left( a + b + c \right)\left( a - b + c \right)} \left[ \because \left( a + b + c \right)\left( a - b + c \right) = a^2 - b^2 + c^2 + 2ac \right]\]

\[ = \frac{\left( a + b + c \right)}{\left( a - b + c \right)} =\text {  RHS }\]