Let `(1)/(x_(1)), (1)/(x_(2)), (1)/(x_(3))`....`(x_(i) ne 0` for i = 1,2,....n)be in A.P.such that `x_(1) = 4` and `x_(21) = 20`.If n is the least positive integer for which `x_(n) gt 50`, then `sum_(i = 1)^(n) ((1)/(x_(i)))` is equal to .

Options

A. 3
B. `(13)/(8)`
C. `(13)/(4)`
D. `(1)/(8)`

Answer

To solve the problem step by step, we will follow the given information and derive the required values systematically.

### Step 1: Identify the first term and the 21st term
We know:
- \( x_1 = 4 \)
- \( x_{21} = 20 \)

### Step 2: Write the general term of the sequence
Since \( \frac{1}{x_1}, \frac{1}{x_2}, \frac{1}{x_3}, \ldots \) are in A.P., we can express the \( n \)-th term of the sequence as:
\[ \frac{1}{x_n} = a + (n-1)d \]
where \( a = \frac{1}{x_1} = \frac{1}{4} \) and \( d \) is the common difference.

### Step 3: Set up the equation for the 21st term
For \( n = 21 \):
\[ \frac{1}{x_{21}} = a + (21-1)d = \frac{1}{4} + 20d \]
We know \( \frac{1}{x_{21}} = \frac{1}{20} \), so we set up the equation:
\[ \frac{1}{20} = \frac{1}{4} + 20d \]

### Step 4: Solve for \( d \)
Rearranging the equation:
\[ 20d = \frac{1}{20} - \frac{1}{4} \]
Finding a common denominator (which is 20):
\[ \frac{1}{4} = \frac{5}{20} \]
Thus,
\[ 20d = \frac{1}{20} - \frac{5}{20} = -\frac{4}{20} = -\frac{1}{5} \]
Now, divide by 20:
\[ d = -\frac{1}{5 \times 20} = -\frac{1}{100} \]

### Step 5: Write the general term for \( \frac{1}{x_n} \)
Now we have:
\[ \frac{1}{x_n} = \frac{1}{4} + (n-1)\left(-\frac{1}{100}\right) \]
This simplifies to:
\[ \frac{1}{x_n} = \frac{1}{4} - \frac{n-1}{100} \]

### Step 6: Solve for \( x_n \)
Rearranging gives:
\[ \frac{1}{x_n} = \frac{25}{100} - \frac{n-1}{100} = \frac{25 - (n-1)}{100} = \frac{26 - n}{100} \]
Thus,
\[ x_n = \frac{100}{26 - n} \]

### Step 7: Find the least positive integer \( n \) such that \( x_n > 50 \)
We need:
\[ \frac{100}{26 - n} > 50 \]
This simplifies to:
\[ 100 > 50(26 - n) \]
\[ 100 > 1300 - 50n \]
Rearranging gives:
\[ 50n > 1200 \implies n > 24 \]
The least positive integer satisfying this is \( n = 25 \).

### Step 8: Calculate the sum \( S_n = \sum_{i=1}^{n} \frac{1}{x_i} \)
Using the formula for the sum of an A.P.:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]
Substituting \( n = 25 \), \( a = \frac{1}{4} \), and \( d = -\frac{1}{100} \):
\[ S_{25} = \frac{25}{2} \left(2 \cdot \frac{1}{4} + (25 - 1)\left(-\frac{1}{100}\right)\right) \]
Calculating:
\[ = \frac{25}{2} \left(\frac{1}{2} - \frac{24}{100}\right) \]
Converting \( \frac{24}{100} \) to a common denominator:
\[ = \frac{25}{2} \left(\frac{50}{100} - \frac{24}{100}\right) = \frac{25}{2} \cdot \frac{26}{100} = \frac{25 \cdot 26}{200} = \frac{650}{200} = \frac{65}{20} = \frac{13}{4} \]

### Final Answer
Thus, the sum \( S_n = \sum_{i=1}^{n} \frac{1}{x_i} \) is:
\[ \frac{13}{4} \]