Question

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

• A. are in A.P.
• B. are in G.P.
• C. are in H.P.
• D. satisfy $ab=cd$

Answer 1

The correct option is B are in G.P.

We have
$(a^2+b^2+c^2)p^2-2(ab+bc+cd)p+(b^2+c^2+d^2)\le 0\cdots \left(1\right)$

$LHS=a^2{p}^2+b^2{p}^2+c^2{p}^2-2abp-2bcp-2cdp+b^2+c^2+d^2$
$=(a^2{p}^2-2abp+b^2)+(b^2{p}^2-2bcp+c^2)+(c^2{p}^2-2cdp+d^2)$
$=(ap-b{)}^2+(bp-c{)}^2+(cp-d{)}^2\ge 0\cdots (2)$
as the sum of squares of real numbers is non-negative.

From equations $\left(1\right)$ and $\left(2\right),$
$(ap-b{)}^2+(bp-c{)}^2+(cp-d{)}^2=0$
This is possible only if
$ap-b=bp-c=cp-d=0$
$\Rightarrow \frac{b}{a}=\frac{c}{b}=\frac{d}{c}=p$
$\therefore a,b,c,d$ are in G.P.