Harmonic Progression Terms and System of Linear Equations Solutions

To solve the problem, we first analyze the given Harmonic Progression (HP) and the system of linear equations.

Step 1: Harmonic Progression Relations

Since $p, q, r$ are the $10^{\text{th}}$, $100^{\text{th}}$, and $1000^{\text{th}}$ terms of a Harmonic Progression, their reciprocals $\frac{1}{p}, \frac{1}{q}, \frac{1}{r}$ form an Arithmetic Progression (AP). Let $a$ be the first term and $d$ be the common difference of this AP. Then:

$$\frac{1}{p} = a + 9d$$ $$\frac{1}{q} = a + 99d$$ $$\frac{1}{r} = a + 999d$$

Step 2: Simplifying the System of Linear Equations

The system of linear equations is:

1. $x + y + z = 1$
2. $10x + 100y + 1000z = 0$
3. $qr x + pr y + pq z = 0$

Dividing equation (3) by $p q r$ (since $p, q, r \neq 0$), we get:

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

Substitute the expressions for $\frac{1}{p}, \frac{1}{q}, \frac{1}{r}$ into this equation:

$$x(a + 9d) + y(a + 99d) + z(a + 999d) = 0$$ $$a(x + y + z) + d(9x + 99y + 999z) = 0 \quad \text{--- (4)}$$

From equations (1) and (2):

• $x + y + z = 1$
• $9x + 99y + 999z = (10x + 100y + 1000z) - (x + y + z) = 0 - 1 = -1$

Substituting these into equation (4):

$$a(1) + d(-1) = 0 \implies a - d = 0 \implies a = d$$

Step 3: Relating $a = d$ to the Ratios of $p, q, r$

Let's check the ratio $\frac{p}{q}$:

$$\frac{p}{q} = \frac{\frac{1}{q}}{\frac{1}{p}} = \frac{a + 99d}{a + 9d}$$

• If $a = d$, then $\frac{p}{q} = \frac{100d}{10d} = 10$.
• Conversely, if $\frac{p}{q} = 10 \implies a + 99d = 10(a + 9d) \implies 9a = 9d \implies a = d$.

Similarly:

$$\frac{q}{r} = 10 \iff a = d$$ $$\frac{p}{r} = 100 \iff a = d$$

Step 4: Analyzing the Solutions of the System

• Case 1: $a = d$ (i.e., $\frac{q}{r} = 10$, $\frac{p}{q} = 10$, or $\frac{p}{r} = 100$)
  Equation (3) is satisfied for all solutions of equations (1) and (2). Thus, the system reduces to 2 independent linear equations in 3 variables, which has infinitely many solutions.

  Let's test if $\left(x = \frac{10}{9}, y = -\frac{1}{9}, z = 0\right)$ is a solution:

  1. $x + y + z = \frac{10}{9} - \frac{1}{9} + 0 = 1$
  2. $10x + 100y + 1000z = 10\left(\frac{10}{9}\right) + 100\left(-\frac{1}{9}\right) + 0 = 0$
  3. $\frac{x}{p} + \frac{y}{q} + \frac{z}{r} = \frac{10}{9}(10d) - \frac{1}{9}(100d) + 0 = 0$

  Therefore, $\left(x = \frac{10}{9}, y = -\frac{1}{9}, z = 0\right)$ is indeed a solution.

• Case 2: $a \neq d$ (i.e., $\frac{p}{r} \neq 100$ or $\frac{p}{q} \neq 10$)
  For any $(x, y, z)$ satisfying equations (1) and (2), the LHS of equation (3) evaluates to $a - d \neq 0$. Thus, equation (3) cannot be satisfied simultaneously, meaning the system has no solution.

Step 5: Matching List-I to List-II

1. (I) If $\frac{q}{r} = 10$, then $a = d$. The system has $x = \frac{10}{9}, y = -\frac{1}{9}, z = 0$ as a solution $\rightarrow$ (Q).
2. (II) If $\frac{p}{r} \neq 100$, then $a \neq d$. The system has no solution $\rightarrow$ (S).
3. (III) If $\frac{p}{q} \neq 10$, then $a \neq d$. The system has no solution $\rightarrow$ (S).
4. (IV) If $\frac{p}{q} = 10$, then $a = d$. The system has infinitely many solutions $\rightarrow$ (R).

Thus, the correct matching is: $(\text{I}) \rightarrow (\text{Q}); \quad (\text{II}) \rightarrow (\text{S}); \quad (\text{III}) \rightarrow (\text{S}); \quad (\text{IV}) \rightarrow (\text{R})$

This corresponds to Option B.