Question:
Let $a, b, c, d$ and $p$ be any non zero distinct real numbers such that $\left(a^{2}+b^{2}+c^{2}\right) p^{2}-2(a b+b c+c d) p+\left(b^{2}+c^{2}+d^{2}\right)$ $=0$. Then :
Correct Option: , 3
Solution:
Rearrange given equation, we get
$\left(a^{2} p^{2}-2 a b p+b^{2}\right)+\left(b^{2} p^{2}-2 b c p+c^{2}\right)$ $+\left(c^{2} p^{2}-2 c d p+d^{2}\right)=0$
$\Rightarrow(a p-b)^{2}+(b p-c)^{2}+(c p-d)^{2}=0$
$\therefore a p-b=b p-c=c p-d=0$
$\Rightarrow \frac{b}{a}=\frac{c}{b}=\frac{d}{c}$ $\therefore a, b, c, d$ are in G.P.