Let `x_(1),x_(2),x_(3), . . .,x_(k)` be the divisors of positive integer '`n`' (including `1` and `x`). If `x_(1)+x_(2)+ . . .+x_(k)=75`, then `sum_(i=1)^(k)(1)/(x_(i))` is equal to:

Options

A. (a) `(k^(2))/(75)`
B. (b) `(75)/(k)`
C. (c) `(n^(2))/(75)`
D. (d) `(75)/(n)`

Explanation

To solve the problem, we need to find the sum of the reciprocals of the divisors of a positive integer \( n \) given that the sum of the divisors is 75. Let's denote the divisors of \( n \) as \( x_1, x_2, \ldots, x_k \).

### Step-by-Step Solution:

1. **Understanding the Divisors**:
The divisors of \( n \) are \( x_1, x_2, \ldots, x_k \) where \( x_1 = 1 \) and \( x_k = n \). The sum of these divisors is given as:
\[ x_1 + x_2 + \ldots + x_k = 75 \]

2. **Reciprocal of the Divisors**:
We need to find the sum:
\[ S = \sum_{i=1}^{k} \frac{1}{x_i} \]

3. **Using the Property of Divisors**:
For each divisor \( x_i \), there is a corresponding divisor \( x_{k+1-i} \) such that:
\[ x_i \cdot x_{k+1-i} = n \]
This means that:
\[ x_{k+1-i} = \frac{n}{x_i} \]

4. **Rewriting the Sum**:
We can rewrite the sum of the reciprocals as:
\[ S = \sum_{i=1}^{k} \frac{1}{x_i} = \sum_{i=1}^{k} \frac{n}{x_i} \cdot \frac{1}{n} \]
This allows us to factor out \( \frac{1}{n} \):
\[ S = \frac{1}{n} \sum_{i=1}^{k} n \cdot \frac{1}{x_i} \]

5. **Identifying the Sum**:
Notice that \( \sum_{i=1}^{k} n \cdot \frac{1}{x_i} \) is equivalent to:
\[ \sum_{i=1}^{k} x_{k+1-i} = x_k + x_{k-1} + \ldots + x_1 \]
which is simply the sum of the divisors:
\[ \sum_{i=1}^{k} x_i = 75 \]

6. **Final Calculation**:
Thus, we have:
\[ S = \frac{1}{n} \cdot 75 \]
Therefore:
\[ S = \frac{75}{n} \]

7. **Conclusion**:
The final result for the sum of the reciprocals of the divisors is:
\[ S = \frac{75}{n} \]

### Answer:
The value of \( \sum_{i=1}^{k} \frac{1}{x_i} \) is \( \frac{75}{n} \).