Let $a, b, c, d$ and $p$ be any non zero distinct real numbers such that $\left(a^{2}+b^{2}+c^{2}\right) p^{2}-2(a b+b c$ $+\mathrm{cd}) \mathrm{p}+\left(\mathrm{b}^{2}+\mathrm{c}^{2}+\mathrm{d}^{2}\right)=0 .$ Then :
Correct Option: , 3
$\left(a^{2}+b^{2}+c^{2}\right) p^{2}+2(a b+b c+c d) p+b^{2}+c^{2}+d^{2}$
$=0$
$\Rightarrow\left(\mathrm{a}^{2} \mathrm{p}^{2}+2 \mathrm{abp}+\mathrm{b}^{2}\right)+\left(\mathrm{b}^{2} \mathrm{p}^{2}+2 \mathrm{bcp}+\mathrm{c}^{2}\right)+$
$\left(\mathrm{c}^{2} \mathrm{p}^{2}+2 \mathrm{cdp}+\mathrm{d}^{2}\right)=0$
$\Rightarrow(a b+b)^{2}+(b p+c)^{2}+(c p+d)^{2}=0$
This is possible only when
$a p+b=0$ and $b p+c=0$ and $c p+d=0$
$\mathrm{p}=-\frac{\mathrm{b}}{\mathrm{a}}=-\frac{\mathrm{c}}{\mathrm{b}}=-\frac{\mathrm{d}}{\mathrm{c}}$
or $\frac{b}{a}=\frac{c}{b}=\frac{d}{c}$
$\therefore a, b, c, d$ are in G.P.
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