Question

If $a,b,c,d\&p$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. $areinarithmeticprogression$
• B. $areingeometricprogression$
• C. $areinharmonicprogression$
• D. $satisfyab=cd$

Answer 1

The correct option is B

$areingeometricprogression$

Step1. Expressing the given data:

Given, $(a^2+b^2+c^2)p^2-2(ab+bc+cd)p+(b^2+c^2+d^2)\le 0,$

$a,b,c,d\&p$are distinct real numbers

Step2. Finding the progression.

Now,

$\begin{array}{rcl}(a^2{p}^2+b^2{p}^2+p^2{c}^2)-2(abp+bcp+cdp)+(b^2+c^2+d^2)& \le & 0\\ (a^2{p}^2+b^2+-2abp)+(b^2{p}^2-2bcp+c^2)+(c^2{p}^2-2cdp+d^2)& \le & 0\\ {(ap-b)}^2+{(bp–c)}^2+{(cp–d)}^2& \le & 0\end{array}$

Sum of the square cannot be negative.

$\begin{array}{rcl}so,(ap-b)& =& 0\\ ap& =& b\dots \left(i\right)\\ (bp–c)& =& 0\\ bp& =& c\dots \left(ii\right)\\ cp–d& =& 0\\ cp& =& d\dots \left(iii\right)\end{array}$

From $\left(i\right)\&\left(ii\right)$

$\begin{array}{rcl}\frac{b}{a}& =& \frac{c}{b}\\ & \Rightarrow & b^2=ac\end{array}$

From $\left(ii\right)\&\left(iii\right)$

$\begin{array}{rcl}\frac{c}{b}& =& \frac{d}{c}\\ & \Rightarrow & c^2=bd\end{array}$

Therefore, $a,b,c\&d$are in Geometric Progression.

Hence correct option is $\mathbf{\left(}\mathit{B}\mathbf{\right)}$