If $a,b,c,d$ and $p$ are distinct real number such that $(a^2+b^2+c^2)p^2-2(ab+bc+dc)p+(b^2+c^2+d^2)\ge 0$, then $a,b,c,d$ :

The correct option is B are in $G.P.$
We have to prove that $=\frac{b}{a}=\frac{c}{d}=\frac{d}{c}$

$\begin{array}{c}Given,\\ \left(a^2+b^2+c^2\right)p^2-2\left(ab+bc+cd\right)p+\left(b^2+c^2+d^2\right)\le 0\\ Proof:\\ a^2{p}^2+b^2{p}^2+c^2{p}^2-2abp-2bcp-2cdp+b^2+c^2+d^2\le 0\\ {\left(ap\right)}^2+b^2-2apbt+{\left(bp\right)}^2+{\left(c\right)}^2-2\times bp\times c+{\left(cp\right)}^2+{\left(d\right)}^2-2\times cp\times d\le 0\\ {\left(ap-b\right)}^2+{\left(bp-c\right)}^2+{\left(cp-d\right)}^2\le 0\\ \therefore {\left(ap-b\right)}^2+{\left(bp-c\right)}^2+{\left(cp-d\right)}^2\le 0\\ \left(ap-b\right)=0,\left(bp-c\right)=0\&\left(cp-d\right)=0\\ =L.H.S\\ \therefore L.H.S=R.H.S\\ So,\left(a^n+b^n\right).\left(b^n+c^n\right)\&\left(c^n+b^n\right)\end{array}$

are in form in $G.P$