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.

Explanation for the correct option:

Simplification of equation:

Solve $(a^2+b^2+c^2)p^2-2(ab+bc+cd)p+(b^2+c^2+d^2)=0$

$$\begin{gathered} \Rightarrow a^2{p}^2+b^2{p}^2+c^2{p}^2-2abp-2bcp-2cdp+b^2+c^2+d^2=0 \\ \Rightarrow (a^2{p}^2-2abp+b^2)+(b^2{p}^2-2bcp+c^2)+(c^2{p}^2-2cdp+d^2)=0 \\ \Rightarrow {(ap-b)}^2+{(bp-c)}^2+{(cp-d)}^2=0 \end{gathered}$$

From this we get,

$ap-b=0=bp-c=cp-d$

$$\begin{gathered} \Rightarrow ap=b\mathrm{or}p=\frac{b}{a} \\ \Rightarrow bp=c\mathrm{or}p=\frac{c}{b} \\ \Rightarrow cp=d\mathrm{or}p=\frac{d}{c} \end{gathered}$$

This means that $p=\frac{b}{a}=\frac{c}{b}=\frac{d}{c}$

Thus, $a,b,c,d$ are in G.P.

Hence, option(B) is correct.