Question

If $a, b, c, d$ and $p$ are distinct real numbers such that $\left(a^{2}+b^{2}+c^{2}\right) p^{2}-2(a b+b c+c d) p+\left(b^{2}+c^{2}+d^{2}\right) \leq 0$ then $a, b, c$ and $d$

• A. are in A.P.
• B. are in G.P.
• C. are in H.P.
• D. satisfy a b=c d

Answer 1

Answer: (B)

We have, $\left(a^{2}+b^{2}+c^{2}\right) p^{2}-2(a b+b c+c d) p+\left(b^{2}+c^{2}+d^{2}\right) \leq 0$
$\Rightarrow (a p-b)^{2}+(b p-c)^{2}+(c p-d)^{2} \leq 0$
$\Rightarrow (a p-b)^{2}+(b p-c)^{2}+(c p-d)^{2}=0$
$(a, b, c, d, p \in R)$
$\Rightarrow a p-b=0,$
$ b p-c=0, c p-d=0$
$\Rightarrow \frac{b}{a}=\frac{c}{b}=\frac{d}{c}=p$
$\Rightarrow a, b, c, d$ are in G.P.