Question

If a, b, c, d and 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 are in

• A. AP
• B. GP
• C. HP
• D. none of these

Answer 1

The correct option is B GP

Consider, $(a^2+b^2+c^2)p^2-2(ab+bc+cd)p+(b^2+c^2+d^2)=0$ ....(1)
Since, $a^2+b^2+c^2>0$ and $(a^2+b^2+c^2)p^2-2(ab+bc+cd)p+(b^2+c^2+d^2)\le 0$

Therefore, roots of the equation (1) are real.
$\Rightarrow {[-2(ab+bc+cd)]}^2-4(a^2+b^2+c^2)(b^2+c^2+d^2)\ge 0$
$\Rightarrow (a^2+b^2+c^2)(b^2+c^2+d^2)\le {(ab+bc+cd)}^2$
$\Rightarrow a^2{b}^2+a^2{c}^2+a^2{d}^2+b^4+b^2{c}^2+b^2{d}^2+c^2{b}^2+c^4+c^2{d}^2\le a^2{b}^2+b^2{c}^2+c^2{d}^2+2ac{b}^2+2bd{c}^2+2abcd$
$\Rightarrow (b^4-2ac{b}^2+a^2{c}^2)+(c^4-2bd{c}^2+b^2{d}^2)+(a^2{d}^2-2abcd+b^2{c}^2)\le 0$
$\Rightarrow {(b^2-ac)}^2+{(c^2-bd)}^2+{(ad-bc)}^2\le 0$
Since, R.H.S of the above inequality cannot be negative.

Therefore, $b^2=ac$; $c^2=bd$; $ad=bc$

and hence, $a,b,c,d$ are in G.P
Ans: B