Prove if $21 (a^{2} + b^{2} + c^{2}) = (a + 2 a b + 4 c)^{2}$ then a,b,c are in G.P.

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Solution

Ans. $21 = 1^{2} + 2^{2} + 4^{2}$ and we know that
$(l_{1}^{2} + m_{1}^{2}) + n_{1}^{2} (l_{2}^{2} + m_{2}^{2} + n_{2}^{2}) - (l_{1} l_{2} + m_{1} m_{2} + n_{2})^{2} = {∣ ∣ ∣ \begin{matrix} l_{1} & m_{1} l_{2} & m_{2} \end{matrix} ∣ ∣ ∣}^{2} + {∣ ∣ ∣ \begin{matrix} m_{1} & n_{1} m_{2} & n_{2} \end{matrix} ∣ ∣ ∣}^{2} + {∣ ∣ ∣ \begin{matrix} n_{1} & l_{1} n_{2} & l_{2} \end{matrix} ∣ ∣ ∣}^{2}$
Above is Lagrange's Identity $(1^{2} + 2^{2} + 4^{2}) (a^{2} + b^{2} + c^{2}) - (1. a + 2. b + 4. c)^{2} = {∣ ∣ ∣ \begin{matrix} 1 & 2 1 & b \end{matrix} ∣ ∣ ∣}^{2} + {∣ ∣ ∣ \begin{matrix} 2 & 4 b & c \end{matrix} ∣ ∣ ∣}^{2} + {∣ ∣ ∣ \begin{matrix} 1 a \end{matrix} ∣ ∣ ∣}^{2}$ or $0 = (b - 2 a)^{2} + 4 (c - 2 b)^{2} + 4 a - c)^{2}$
Hence each bracket is zero.
$∴ b = 2 a, 2 b = c, c = 4 a$ or $a = \frac{b}{2} = \frac{c}{4} = k ∴ a, b, c$ are $k, 2 k, 4 k$ which are in G.P of common ratio 2.

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