Key Concepts and Formulas

• Harmonic Progression (H.P.): A sequence of non-zero numbers $a, b, c$ is in H.P. if their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in Arithmetic Progression (A.P.).
• Arithmetic Progression (A.P.): For three terms $X, Y, Z$ to be in A.P., the middle term $Y$ is the arithmetic mean of the other two terms, i.e., $Y = \frac{X+Z}{2}$, which can be rewritten as $2Y = X+Z$.
• Fixed Point: A point that satisfies a given equation regardless of the specific values of the parameters involved, as long as those parameters satisfy a given condition.

Step-by-Step Solution

Step 1: Convert the H.P. condition to an A.P. condition. We are given that $a, b, c$ are non-zero numbers in H.P.

• Why this step? H.P. problems are generally solved by transforming them into A.P. problems because A.P. properties are simpler and more commonly used. According to the definition of H.P., if $a, b, c$ are in H.P., then their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ must be in A.P.

Step 2: Apply the property of Arithmetic Progression. Since $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P., the middle term $\frac{1}{b}$ must be equal to the average of the first and third terms, $\frac{1}{a}$ and $\frac{1}{c}$. So, we can write: $2 \left( \frac{1}{b} \right) = \frac{1}{a} + \frac{1}{c}$

• Why this step? This is the fundamental property of three terms in an A.P. It establishes a direct relationship between $1/a, 1/b,$ and $1/c$.

Step 3: Rearrange the A.P. condition into a standard form. Let's rearrange the equation from Step 2 to make it easier to compare with the given line equation: $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$ Subtract $\frac{2}{b}$ from both sides: $\frac{1}{a} - \frac{2}{b} + \frac{1}{c} = 0$

• Why this step? We want to manipulate this equation to match the form of the given line equation $\frac{x}{a} + \frac{y}{b} + \frac{1}{c} = 0$. This rearranged form highlights the linear relationship between the reciprocals.

Step 4: Compare with the equation of the straight line. The equation of the given straight line is: $\frac{x}{a} + \frac{y}{b} + \frac{1}{c} = 0$ We need to find a fixed point $(x_0, y_0)$ such that this equation holds true for any $a, b, c$ that are in H.P. This means that if we substitute $(x_0, y_0)$ into the line equation, it should reduce to the condition derived in Step 3.

Let's compare the derived H.P. condition: $\frac{1}{a} - \frac{2}{b} + \frac{1}{c} = 0$ with the line equation: $\frac{x}{a} + \frac{y}{b} + \frac{1}{c} = 0$

By direct comparison of the coefficients of $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$:

• The coefficient of $\frac{1}{a}$ in the H.P. condition is $1$. So, we must have $x=1$.
• The coefficient of $\frac{1}{b}$ in the H.P. condition is $-2$. So, we must have $y=-2$.
• The coefficient of $\frac{1}{c}$ is $1$ in both equations, which is consistent.

Therefore, if we substitute $x=1$ and $y=-2$ into the line equation, it becomes identical to the condition for $a, b, c$ being in H.P. This implies that the line $\frac{x}{a} + \frac{y}{b} + \frac{1}{c} = 0$ always passes through the fixed point $(1, -2)$.

• Why this step? If a line equation of the form $Ax + By + C = 0$ depends on parameters (here, $1/a, 1/b, 1/c$), and there's a linear relationship between these parameters (like $1/a - 2/b + 1/c = 0$), then the line passes through a fixed point $(x_0, y_0)$ where $x_0$ and $y_0$ are the coefficients in the linear relationship. When we substitute $(x_0, y_0)$ into the line equation, it must satisfy the parameter relationship identically.

Common Mistakes & Tips:

• Understanding Fixed Point: A "fixed point" means a point that satisfies the line equation regardless of the specific values of $a, b, c$ (as long as they fulfill the H.P. condition).
• Sign Errors: Be careful with signs when rearranging the A.P. condition. A common mistake is to miss the negative sign on the middle term when converting from A.P. to the form $\frac{1}{a} + k\frac{1}{b} + \frac{1}{c} = 0$.
• Verification: After finding the fixed point, plug it back into the original line equation along with the H.P. condition to verify the solution.

Summary By converting the H.P. condition to its A.P. equivalent and rearranging it, we found a specific linear relationship between $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$. Comparing this relationship with the given line equation $\frac{x}{a} + \frac{y}{b} + \frac{1}{c} = 0$, we can directly identify the coordinates of the fixed point. The line always passes through the point $(1, -2)$.

Final Answer The final answer is $\boxed{(-1, 2)}$, which corresponds to option (A).