If `H_1.,H_2,…,H_20` are 20 harmonic means between 2 and 3, then `(H_1+2)/(H_1-2)+(H_20+3)/(H_20-3)=`

Options

A. 20
B. 21
C. 40
D. 38

Explanation

To solve the problem, we need to find the value of \((H_1 + 2)/(H_1 - 2) + (H_{20} + 3)/(H_{20} - 3)\), where \(H_1, H_2, \ldots, H_{20}\) are the 20 harmonic means between 2 and 3.

### Step 1: Understanding Harmonic Means
The harmonic means between two numbers \(a\) and \(b\) can be found using the formula:
\[ H_n = \frac{2ab}{a + b + (n-1)d} \]
where \(d\) is the common difference in the corresponding arithmetic progression (AP) of the reciprocals of the harmonic means.

### Step 2: Finding the Common Difference \(d\)
Given \(a = 2\) and \(b = 3\), we first calculate the common difference \(d\):
\[ d = \frac{b - a}{n + 1} = \frac{3 - 2}{20 + 1} = \frac{1}{21} \]

### Step 3: Finding \(H_1\) and \(H_{20}\)
Using the formula for harmonic means:
- For \(H_1\):
\[ H_1 = \frac{2 \cdot 2 \cdot 3}{2 + 3 + (1-1)d} = \frac{12}{5} = 2.4 \]

- For \(H_{20}\):
\[ H_{20} = \frac{2 \cdot 2 \cdot 3}{2 + 3 + (20-1)d} = \frac{12}{5 + 19 \cdot \frac{1}{21}} = \frac{12}{5 + \frac{19}{21}} = \frac{12}{\frac{105 + 19}{21}} = \frac{12 \cdot 21}{124} = \frac{252}{124} = \frac{63}{31} \approx 2.0323 \]

### Step 4: Substitute \(H_1\) and \(H_{20}\) into the Expression
Now we substitute \(H_1\) and \(H_{20}\) into the expression:
\[ \frac{H_1 + 2}{H_1 - 2} + \frac{H_{20} + 3}{H_{20} - 3} \]

Calculating each term:
1. For \(H_1 = 2.4\):
\[ \frac{H_1 + 2}{H_1 - 2} = \frac{2.4 + 2}{2.4 - 2} = \frac{4.4}{0.4} = 11 \]

2. For \(H_{20} \approx 2.0323\):
\[ \frac{H_{20} + 3}{H_{20} - 3} = \frac{2.0323 + 3}{2.0323 - 3} = \frac{5.0323}{-0.9677} \approx -5.2 \]

### Step 5: Combine the Results
Now, we add the two results:
\[ 11 - 5.2 = 5.8 \]

### Final Answer
Thus, the final answer is:
\[ \frac{H_1 + 2}{H_1 - 2} + \frac{H_{20} + 3}{H_{20} - 3} \approx 5.8 \]