Question

If $a,b,c,d$ and $p$ are distinct 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 AP
• B. are in GP
• C. are in HP
• D. satisfy ab = cd

Answer 1

Answer: (B)

Here, $(a^2+b^2+c^2)p^2-2(ab+bc+cd)p$
$+(b^2+c^2+d^2)\le 0,$
$\Rightarrow (a^2{p}^2-2abp+b^2)+(b^2{p}^2-2bcp+c^2)$
$+(c^2{p}^2-2cdp+d^2)\le 0$
$\Rightarrow (ap-b{)}^2+(bp-c{)}^2+(cp-d{)}^2\le 0$
[since, sum of squares is never less than zero]
Since, each of the squares is zero.
$\therefore (ap-b{)}^2=(bp-c{)}^2=(cp-d{)}^2=0$
$\Rightarrow p=\frac{b}{a}=\frac{c}{d}=\frac{d}{c}$
$\therefore a,b,c,d$ are in GP.