If a, b, c and p are distinct real numbers such that $\left(a^2+b^2+c^2\right)p^2-2(ab+bc+cd)p$$+\left(b^2+c^2+d^2\right)=0$ then a, b, c, d

Given : $\left({\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2\right){\mathrm{p}}^2-2(\mathrm{ab}+\mathrm{bc}+\mathrm{cd})\mathrm{p}+\left({\mathrm{b}}^2+{\mathrm{c}}^2+{\mathrm{d}}^2\right)=0$

$\Rightarrow \left({\mathrm{a}}^2{\mathrm{p}}^2-2\mathrm{abp}+{\mathrm{b}}^2\right)+\left({\mathrm{b}}^2{\mathrm{p}}^2-2\mathrm{bcp}+{\mathrm{c}}^2\right)+\left({\mathrm{c}}^2{\mathrm{p}}^2-2\mathrm{cdp}+{\mathrm{d}}^2\right)=0$

$\Rightarrow (ap-b{)}^2+(bp-c{)}^2+(cp-d{)}^2=0$

Sum of three perfect squares is zero 

$\begin{array}{l}\therefore \mathrm{ap}-\mathrm{b}=\mathrm{bp}-\mathrm{c}=\mathrm{cp}-\mathrm{d}=0\\ \Rightarrow \mathrm{p}=\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{b}}=\frac{\mathrm{d}}{\mathrm{c}}\\ \Rightarrow \mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\text{ are in G.P. }\end{array}$