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(B) Since $p, q, r$ are the $10^{	ext{th}}, 100^{	ext{th}}, 1000^{	ext{th}}$ terms of a harmonic progression,their reciprocals $\frac{1}{p}, \frac{

Vedclass · Accepted Answer

$(I) \rightarrow (Q); (II) \rightarrow (S); (III) \rightarrow (S); (IV) \rightarrow (R)$

Vedclass · Answer

(B) Since $p, q, r$ are the $10^{	ext{th}}, 100^{	ext{th}}, 1000^{	ext{th}}$ terms of a harmonic progression,their reciprocals $\frac{1}{p}, \frac{1}{q}, \frac{1}{r}$ are in arithmetic progression. Let the $AP$ be $a + (n-1)d$. Then $\frac{1}{p} = a + 9d, \frac{1}{q} = a + 99d, \frac{1}{r} = a + 999d$.Dividing the third equation $qrx + pry + pqz = 0$ by $pqr$,we get $\frac{x}{p} + \frac{y}{q} + \frac{z}{r} = 0$.Substituting the $AP$ terms,the system becomes:$x+y+z=1$$x+10y+100z=0$$\frac{x}{p} + \frac{y}{q} + \frac{z}{r} = 0$Calculating the determinant $D$ of the system: $D = \begin{vmatrix} 1 & 1 & 1 \ 1 & 10 & 100 \ \frac{1}{p} & \frac{1}{q} & \frac{1}{r} \end{vmatrix}$.Using properties of $AP$,$D = 0$ if and only if the ratios of the terms match the progression. Specifically,the system has infinitely many solutions if $D=D_x=D_y=D_z=0$. For $(I)$,if $\frac{q}{r}=10$,the system is consistent with infinitely many solutions $(R)$,and since it has infinitely many solutions,it has at least one solution $(T)$.For $(II)$,if $\frac{p}{r} 
eq 100$,the system is inconsistent,leading to no solution $(S)$.For $(III)$,if $\frac{p}{q} 
eq 10$,the system is inconsistent,leading to no solution $(S)$.For $(IV)$,if $\frac{p}{q}=10$,the system has infinitely many solutions $(R)$,and thus at least one solution $(T)$.Matching these,$(I) ightarrow (R, T), (II) ightarrow (S), (III) ightarrow (S), (IV) ightarrow (R, T)$. Option $(B)$ is the correct match.

$List-I$	$List-II$
$(I)$ If $\frac{q}{r}=10$,then the system of linear equations has	$(P)$ $x=0, y=\frac{10}{9}, z=-\frac{1}{9}$ as a solution
$(II)$ If $\frac{p}{r} \neq 100$,then the system of linear equations has	$(Q)$ $x=\frac{10}{9}, y=-\frac{1}{9}, z=0$ as a solution
$(III)$ If $\frac{p}{q} \neq 10$,then the system of linear equations has	$(R)$ infinitely many solutions
$(IV)$ If $\frac{p}{q}=10$,then the system of linear equations has	$(S)$ no solution
	$(T)$ at least one solution

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