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STD 11 - 8. sequence and series — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 8. sequence and series4 Marks
MCQ
Let $a, b, c, d$ and $p$ be any non zero distinct real numbers such that  $\left(a^{2}+b^{2}+c^{2}\right) p^{2}-2(a b+b c+ cd ) p +\left( b ^{2}+ c ^{2}+ d ^{2}\right)=0 .$ Then
  • A
    $a,c,p$ are in $G.P.$ 
  • B
    $a,c,p$ are in $A.P.$ 
  • ✓
    $a,b,c,d$ are in $G.P.$ 
  • D
    $a,b,c,d$ are in $A.P.$
✓

Answer

Correct option: C.
$a,b,c,d$ are in $G.P.$ 
c
$\left(a^{2}+b^{2}+c^{2}\right) p^{2}+2(a b+b c+c d) p+b^{2}+c^{2}+d^{2}$

$=0$

$\Rightarrow\left(a^{2} p^{2}+2 a b p+b^{2}\right)+\left(b^{2} p^{2}+2 b c p+c^{2}\right)+$

$\left(c^{2} p^{2}+2 c d p+d^{2}\right)=0$

$\Rightarrow(a b+b)^{2}+(b p+c)^{2}+(c p+d)^{2}=0$

This is possible only when $a p+b=0$ and $b p+c=0$ and $c p+d=0$

$p =-\frac{ b }{ a }=-\frac{ c }{ b }=-\frac{ d }{ c }$

or $\frac{ b }{ a }=\frac{ c }{ b }=\frac{ d }{ c }$

$\therefore a , b , c , d$ are in $G . P$

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