Let p, q, r be nonzero real numbers that are, respectively, the 10^th, 100^th and 1000^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
(I) If $\frac{r}{q}=10$, then the system of linear equations has	(P) $x=0,y=\frac{10}{9},z=-\frac{1}{9}$ as a solution
(II) If $\frac{r}{p}\ne 100$, then the system of linear equations has	(Q) $x=\frac{10}{9},y=-\frac{1}{9},z=0$ as a solution
(III) If $\frac{q}{p}\ne 10$, then the system of linear equations has	(R) infinitely many solutions
(IV) If $\frac{q}{p}=10$, then the system of linear equations has	(S) no solution
	(T) at least one solution

The correct option is :

The correct option is (I) → (Q); (II) → (S); (III) → (S); (IV) → (R).

Step-by-step Explanation:

Let $p$, $q$, and $r$ be the 10^th, 100^th, and 1000^th terms of a harmonic progression (HP).
This means that their reciprocals $\frac{1}{p}$, $\frac{1}{q}$, and $\frac{1}{r}$ are the 10^th, 100^th, and 1000^th terms of an arithmetic progression (AP).
Let $a$ be the first term and $d$ be the common difference of this AP. Then we can write:

$\frac{1}{p}=a+9d$

$\frac{1}{q}=a+99d$

$\frac{1}{r}=a+999d$

Now, let's analyze the given system of linear equations:

1) $x+y+z=1$
2) $10x+100y+1000z=0$
3) $qrx+pry+pqz=0$

Since $p,q,r$ are non-zero, we can divide the third equation by $pqr$ to rewrite it as:

$\frac{x}{p}+\frac{y}{q}+\frac{z}{r}=0$

Substitute the terms of the AP into this equation:

$x(a+9d)+y(a+99d)+z(a+999d)=0$

Rearranging the terms:
$a(x+y+z)+d(9x+99y+999z)=0$

From the first equation, we know that $x+y+z=1$. This simplifies our equation to:

$a+d(9x+99y+999z)=0$

From the second equation, we have:

$10x+100y+1000z=0$

which can be written as:
$(9x+99y+999z)+(x+y+z)=0$

Since $x+y+z=1$, we obtain:
$9x+99y+999z=-1$

Substituting this result back, we get:

$a-d=0\Rightarrow a=d$

Thus, we have:

$\frac{1}{p}=10d$, $\frac{1}{q}=100d$, and $\frac{1}{r}=1000d$

This gives:
$\frac{r}{q}=\frac{100d}{1000d}=10$
and
$\frac{q}{p}=\frac{10d}{100d}=10$

Let's analyze the conditions in List-I:

Case (I): If $\frac{r}{q}=10$
This matches the derived relationship when the equations are consistent. Let's check if $x=\frac{10}{9},y=-\frac{1}{9},z=0$ (Option Q) is a solution:
1) $x+y+z=\frac{10}{9}-\frac{1}{9}+0=1$ (True)
2) $10x+100y+1000z=10\left(\frac{10}{9}\right)+100(-\frac{1}{9})=0$ (True)
3) $\frac{x}{p}+\frac{y}{q}+\frac{z}{r}=\frac{10}{9p}-\frac{1}{9q}=0\Rightarrow \frac{q}{p}=10$, which is consistent.
So, (I) → (Q).

Case (II): If $\frac{r}{p}\ne 100$
Since $\frac{1}{p}=10d$ and $\frac{1}{r}=1000d$, we must have $\frac{r}{p}=100$ for the harmonic progression equations to hold under consistent parameters. If $\frac{r}{p}\ne 100$, the system becomes inconsistent, leading to no solution.
So, (II) → (S).

Case (III): If $\frac{q}{p}\ne 10$
Similarly, a consistent progression requires $\frac{q}{p}=10$. If $\frac{q}{p}\ne 10$, the system has no solution.
So, (III) → (S).

Case (IV): If $\frac{q}{p}=10$
This satisfies the consistency condition $a=d$, which makes the determinant of the coefficient matrix zero and the system consistent. Therefore, there are infinitely many solutions.
So, (IV) → (R).