Question

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

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

Answer 1

The correct option is B $a,b,c,d$ are in G.P.

$(a^2+b^2+c^2)p^2-2(ab+bc+cd)p+(b^2+c^2+d^2)=0$
$(a^2{p}^2-2abp+b^2)+[b^2{p}^2-2bcp+c^2]+[c^2{p}^2-2cdp+d^2]=0$
$(ap-b{)}^2+(bp-c{)}^2+(cp-d{)}^2=0$
$ap=b$
$bp=c$
$cp=d$
$\frac{b}{a}=\frac{c}{b}=\frac{d}{c}=p$
Thus $a,b,c,d$ are in G.P