Question

Let $a,b,c,d$ and $p$ be any non zero distinct real numbers such that $\left(a^2+b^2+c^2\right)p^2-2(ab+bc$ $+cd)p+\left(b^2+c^2+d^2\right)=0.$ Then

• A. $a,c,p$ are in $G.P.$
• B. $a,c,p$ are in $A.P.$
• C. $a,b,c,d$ are in $G.P.$
• D. $a,b,c,d$ are in $A.P.$

Answer 1

Answer: (C)

$\left(a^2+b^2+c^2\right)p^2+2(ab+bc+cd)p+b^2+c^2+d^2$
$=0$
$\Rightarrow \left(a^2{p}^2+2abp+b^2\right)+\left(b^2{p}^2+2bcp+c^2\right)+\left(c^2{p}^2+2cdp+d^2\right)=0$
$\Rightarrow (ab+b{)}^2+(bp+c{)}^2+(cp+d{)}^2=0$
This is possible only when
$ap+b=0$ and $bp+c=0$ and $cp+d=0$
$p=-\frac{b}{a}=-\frac{c}{b}=-\frac{d}{c}$
or $\frac{b}{a}=\frac{c}{b}=\frac{d}{c}$
$\therefore a,b,c,d$ are in G.P.