Q.

Let p,q,r  be non-zero real numbers that are respectively the 10th,100th  and  1000th   terms of a harmonic progression. Consider the system of linear equations:
x+y+z=110x+100y+1000z=0 , (qr)x+(pr)y+(pq)z=0  

 Column-I Column-II
(I)If  qr=10,  then the system of linear equations has(P)x=0,y=109,z=19  as a solution
(II)If pr100,  then the system of linear equations has(Q)x=109,y=19,z=0  as a solution
(III)If pq10,  then the system of linear equations has(R)infinitely many solutions
(IV)If  pq=10,  then the system of linear equations has(S)no solution
  (T)at least one solution

The correct option is:

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a

(I)-(T); (II)-(S); (III)-(P); (IV)-(T)

b

(I)-(Q); (II)-(R); (III)-(P); (IV)-(R) 

c

(I)-(T); (II)-(R); (III)-(S); (IV)-(T) 

d

(I)-(Q); (II)-(S); (III)-(S); (IV)-(R)

answer is B.

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Detailed Solution

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(I) qr=10D=Dx=Dy=Dz=0 
So, there will be infinitely many solutions and they will be of the type (x,y,z)=(109+10λ,1911λ,λ)  which gives (I)-(Q),(R),(T)
(II)  pr100Dy0
So, no solution which gives (II)-(S)
(III)  pq10Dz0
So, no solution which gives (III)-(S)
(IV)  pq=10Dx=Dy=Dz=D=0
So, infinitely many solutions which give (IV)-(Q),(R),(T)

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