Question

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

Given,

$$\begin{gathered} \left(a^2+b^2+c^2\right)p^2-2\left(ab+bc+cd\right)p+\left(b^2+c^2+d^2\right)=0 \\ \Rightarrow a^2{p}^2+b^2{p}^2+c^2{p}^2-2pab-2pbc-2pcd+b^2+c^2+d^2=0 \\ \Rightarrow \left(a^2{p}^2-2pab+b^2\right)+\left(b^2{p}^2-2pbc+c^2\right)+\left(c^2{p}^2-2pcd+d^2\right)=0\left[\because {\left(a-b\right)}^2=a^2-2ab-b^2\right] \\ \Rightarrow {\left(ap-b\right)}^2+{\left(bp-c\right)}^2+{\left(cp-d\right)}^2=0\dots \dots \left(i\right) \end{gathered}$$

$a,b,c,d$ and $p$ be any non zero distinct real numbers

Therefore, from equation $\left(i\right)$, we get

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

Geometric Progression (G.P) is a type of sequence where each succeeding term is obtained by multiplying the common ratio by the preceding term.

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

Hence, option (B) is the correct answer.