If $a^{2} + b^{2}, a b + b c, b^{2} + c^{2}$ are in G.P. then prove that a,b,c are also in G.P.

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Solution

$a^{2} + b^{2}, a b + b c, b^{2} + c^{2}$ are in $G P$

$(a b + b c)^{2} = (a^{2} + b^{2}) (b^{2} + c^{2})$

$a^{2} b^{2} + b^{2} c^{2} + 2 a b^{2} c = a^{2} b^{2} + a^{2} c^{2} + b^{2} c^{2} + b^{4}$

$b^{4} + a^{2} c^{2} - 2 a b^{2} c = 0$

$(b^{2} - a c)^{2} = 0$

$b^{2} = a c$

$\Rightarrow a, b, c$ are in $G P$

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