Question

Let $p,q,r$ be nonzero real numbers that are, respectively, the $10^{\text{th }},100^{\text{th }}$ and $1000^{\text{th }}$ terms of a harmonic progression. Consider the system of linear equations
$x+y+z=1$
$10x+100y+1000z=0$
$qrx+pry+pqz=0.$

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

The correct option is:

• A. (I) $\to$ (T); (II) $\to$ (R); (III) $\to$ (S); (IV) $\to$ (T)
• B. (I) $\to$ (Q); (II) $\to$ (S); (III) $\to$ (S); (IV) $\to$ (R)
• C. (I) $\to$ (Q); (II) $\to$ (R); (III) $\to$ (P); (IV) $\to$ (R)
• D. (I) $\to$ (T); (II) $\to$ (S); (III) $\to$ (P); (IV) $\to$ (T)

Answer 1

Answer: (B)

If $\frac{q}{r}=10\Rightarrow A=D\Rightarrow D_x=D_y=D_z=0$
So, there are infinitely many solutions Look of infinitely many solutions can be given as
$x+y+z=1$
$\&10x+100y+1000z=0\Rightarrow x+10y+100z=0$
Let $z=\lambda$
then $x+y=1-\lambda$
and $x+10y=-100\lambda$
$\Rightarrow x=\frac{10}{9}+10\lambda ;y=\frac{-1}{9}-11\lambda$
i.e., $(x,y,z)\equiv \left(\frac{10}{9}+10\lambda ,\frac{-1}{9}-11\lambda ,\lambda \right)$
$Q\left(\frac{10}{9},\frac{-1}{9},0\right)$ valid for $\lambda =0$
$P\left(0,\frac{10}{9},\frac{-1}{9}\right)$ not valid for any $\lambda$.
(I) $\to$ Q,R,T
(II) If $\frac{p}{r}\ne 100$, then $D_y\ne 0$
So no solution
(II) $\to$ (S)
(III) If $\frac{p}{q}\ne 10$, then $D_2\ne 0$ so, no solution
(III) $\to$ (S)
(IV) If $\frac{p}{q}=10\Rightarrow D_z=0\Rightarrow D_x=D_y=0$
so infinitely many solution
(IV) $\to$ Q, R, T