If `H_1. H_2...., H_n` are n harmonic means between a and b`(!=a)`, then the value of `(H_1+a)/(H_1-a)+(H_n+b)/(H_n-b)`=

Options

A. 0
B. n
C. 2n
D. 1

Answer

To solve the problem, we need to find the value of the expression:

\[ \frac{H_1 + a}{H_1 - a} + \frac{H_n + b}{H_n - b} \]

where \(H_1, H_2, \ldots, H_n\) are \(n\) harmonic means between \(a\) and \(b\).

### Step 1: Understanding Harmonic Means
Since \(H_1, H_2, \ldots, H_n\) are harmonic means between \(a\) and \(b\), we know that the reciprocals \( \frac{1}{H_1}, \frac{1}{H_2}, \ldots, \frac{1}{H_n}, \frac{1}{b} \) form an arithmetic progression (AP) with \( \frac{1}{a} \) as the first term.

### Step 2: Finding the Common Difference
Let the common difference of the AP be \(d\). Then we can express the harmonic means as:
\[ \frac{1}{H_1} = \frac{1}{a} + d, \quad \frac{1}{H_n} = \frac{1}{b} - d \]

### Step 3: Expressing \(H_1\) and \(H_n\)
From the above equations, we can express \(H_1\) and \(H_n\) as:
\[ H_1 = \frac{1}{\frac{1}{a} + d}, \quad H_n = \frac{1}{\frac{1}{b} - d} \]

### Step 4: Substituting into the Expression
Now we substitute \(H_1\) and \(H_n\) into the expression:
\[ \frac{H_1 + a}{H_1 - a} + \frac{H_n + b}{H_n - b} \]

### Step 5: Simplifying Each Term
1. For the first term:
\[ \frac{H_1 + a}{H_1 - a} = \frac{\frac{1}{\frac{1}{a} + d} + a}{\frac{1}{\frac{1}{a} + d} - a} \]
Simplifying this gives:
\[ = \frac{\frac{1 + a(\frac{1}{a} + d)}{(\frac{1}{a} + d)}}{\frac{1 - a(\frac{1}{a} + d)}{(\frac{1}{a} + d)}} = \frac{1 + 1 + ad}{1 - 1 - ad} = \frac{2 + ad}{-ad} \]

2. For the second term:
\[ \frac{H_n + b}{H_n - b} = \frac{\frac{1}{\frac{1}{b} - d} + b}{\frac{1}{\frac{1}{b} - d} - b} \]
Simplifying this gives:
\[ = \frac{\frac{1 + b(\frac{1}{b} - d)}{(\frac{1}{b} - d)}}{\frac{1 - b(\frac{1}{b} - d)}{(\frac{1}{b} - d)}} = \frac{1 + 1 - bd}{1 - 1 + bd} = \frac{2 - bd}{bd} \]

### Step 6: Combining the Results
Now we combine the two simplified terms:
\[ \frac{2 + ad}{-ad} + \frac{2 - bd}{bd} \]

### Step 7: Finding a Common Denominator
The common denominator is \(-abd\):
\[ = \frac{(2 + ad)bd - (2 - bd)ad}{-abd} \]

### Step 8: Final Simplification
After simplifying, we find that the expression reduces to:
\[ \frac{(2bd + a^2d - 2ad + b^2d)}{-abd} \]

### Final Answer
Thus, the final value of the expression is:
\[ \frac{2(b - a)}{(b - a)(-d)} = -2 \]