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Q.
Let $a , b , c , d$ and $p$ be any non zero distinct real numbers such that $\left(a^{2}+b^{2}+c^{2}\right) p^{2}-2(a b+b c$ $+ cd ) p +\left( b ^{2}+ c ^{2}+ d ^{2}\right)=0 .$ Then
$\left(a^{2}+b^{2}+c^{2}\right) p^{2}+2(a b+b c+c d) p+b^{2}+c^{2}+d^{2}$
$=0$
$\Rightarrow \left(a^{2} p^{2}+2 a b p+b^{2}\right)+\left(b^{2} p^{2}+2 b c p+c^{2}\right)+ \left(c^{2} p^{2}+2 c d p+d^{2}\right)=0$
$\Rightarrow (a b+b)^{2}+(b p+c)^{2}+(c p+d)^{2}=0$
This is possible only when
$ap +b=0$ and $b p+c=0$ and $c p+d=0$
$p =-\frac{ b }{ a }=-\frac{ c }{ b }=-\frac{ d }{ c }$
or $\frac{b}{a}=\frac{c}{b}=\frac{d}{c}$
$\therefore a , b , c , d$ are in G.P.