In a harmonic progression `t_(1), t_(2), t_(3),…………….,` it is given that `t_(5)=20 and t_(6)=50`. If `S_(n)` denotes the sum of first n terms of this, then the value of n for which `S_(n)` is maximum is

Options

A. 6
B. 7
C. 9
D. 10

Answer

To solve the problem step by step, we will analyze the given harmonic progression (HP) and derive the necessary values to find the maximum sum \( S_n \).

### Step 1: Understand the relationship between HP and AP
In a harmonic progression, the terms can be expressed as the reciprocals of an arithmetic progression (AP). Thus, if \( t_1, t_2, t_3, \ldots \) are the terms of the HP, then \( \frac{1}{t_1}, \frac{1}{t_2}, \frac{1}{t_3}, \ldots \) will be in AP.

### Step 2: Define the terms in AP
Let the first term of the corresponding AP be \( A \) and the common difference be \( D \). Therefore, the \( n \)-th term of the AP can be expressed as:
\[ \frac{1}{t_n} = A + (n-1)D \]

### Step 3: Use the given terms
We know:
- \( t_5 = 20 \) implies \( \frac{1}{t_5} = \frac{1}{20} \)
- \( t_6 = 50 \) implies \( \frac{1}{t_6} = \frac{1}{50} \)

From the AP representation:
\[ A + 4D = \frac{1}{20} \quad \text{(1)} \]
\[ A + 5D = \frac{1}{50} \quad \text{(2)} \]

### Step 4: Solve the equations
Subtract equation (1) from equation (2):
\[ (A + 5D) - (A + 4D) = \frac{1}{50} - \frac{1}{20} \]
This simplifies to:
\[ D = \frac{1}{50} - \frac{1}{20} \]
Finding a common denominator (100):
\[ D = \frac{2 - 5}{100} = -\frac{3}{100} \]

### Step 5: Substitute \( D \) back to find \( A \)
Substituting \( D \) into equation (1):
\[ A + 4\left(-\frac{3}{100}\right) = \frac{1}{20} \]
\[ A - \frac{12}{100} = \frac{5}{100} \]
\[ A = \frac{5}{100} + \frac{12}{100} = \frac{17}{100} \]

### Step 6: Find the maximum \( S_n \)
The sum of the first \( n \) terms of the HP, denoted \( S_n \), can be expressed as:
\[ S_n = t_1 + t_2 + \ldots + t_n \]
Since \( S_n \) is maximum when the last term of the corresponding AP is positive, we need to find \( n \) such that:
\[ A + (n-1)D > 0 \]
Substituting \( A \) and \( D \):
\[ \frac{17}{100} + (n-1)\left(-\frac{3}{100}\right) > 0 \]
Multiplying through by 100 to eliminate the fraction:
\[ 17 - 3(n-1) > 0 \]
This simplifies to:
\[ 17 - 3n + 3 > 0 \]
\[ 20 > 3n \]
\[ n < \frac{20}{3} \approx 6.67 \]

### Step 7: Determine the maximum integer \( n \)
Since \( n \) must be an integer, the maximum value of \( n \) is:
\[ n = 6 \]

### Conclusion
Thus, the value of \( n \) for which \( S_n \) is maximum is:
\[ \boxed{6} \]