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, we need to find the value of \( n \) for which the sum \( S_n \) of the first \( n \) terms of a harmonic progression (HP) is maximum. We are given that \( t_5 = 20 \) and \( t_6 = 50 \).

### Step 1: Understanding the Harmonic Progression
In a harmonic progression, the terms can be expressed as the reciprocals of an arithmetic progression (AP). If the \( n \)-th term of the HP is given by \( t_n \), then we can express it as:
\[ t_n = \frac{1}{a + (n-1)d} \]
where \( a \) and \( d \) are constants.

### Step 2: Setting Up the Equations
From the problem, we know:
- \( t_5 = 20 \) implies:
\[ \frac{1}{a + 4d} = 20 \quad \Rightarrow \quad a + 4d = \frac{1}{20} \quad \text{(Equation 1)} \]
- \( t_6 = 50 \) implies:
\[ \frac{1}{a + 5d} = 50 \quad \Rightarrow \quad a + 5d = \frac{1}{50} \quad \text{(Equation 2)} \]

### Step 3: Solving the Equations
Now we have two equations:
1. \( a + 4d = \frac{1}{20} \)
2. \( a + 5d = \frac{1}{50} \)

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 (which is 100):
\[ d = \frac{2}{100} - \frac{5}{100} = -\frac{3}{100} \]

### Step 4: Finding \( a \)
Substituting \( d \) back into Equation 1:
\[ a + 4\left(-\frac{3}{100}\right) = \frac{1}{20} \]
\[ a - \frac{12}{100} = \frac{1}{20} \]
Converting \( \frac{1}{20} \) to a fraction with a denominator of 100:
\[ \frac{1}{20} = \frac{5}{100} \]
Thus, we have:
\[ a - \frac{12}{100} = \frac{5}{100} \]
\[ a = \frac{5}{100} + \frac{12}{100} = \frac{17}{100} \]

### Step 5: General Term of the HP
Now we can express the \( n \)-th term of the HP:
\[ t_n = \frac{1}{a + (n-1)d} = \frac{1}{\frac{17}{100} + (n-1)\left(-\frac{3}{100}\right)} = \frac{100}{17 - 3n + 3} = \frac{100}{20 - 3n} \]

### Step 6: Finding Maximum \( S_n \)
The sum of the first \( n \) terms \( S_n \) is:
\[ S_n = t_1 + t_2 + t_3 + \ldots + t_n \]
To find when \( S_n \) is maximum, we need to determine when \( t_n < 0 \):
\[ \frac{100}{20 - 3n} < 0 \quad \Rightarrow \quad 20 - 3n < 0 \quad \Rightarrow \quad 20 < 3n \quad \Rightarrow \quad n > \frac{20}{3} \approx 6.67 \]
Thus, the first integer \( n \) that makes \( t_n < 0 \) is \( n = 7 \).

### Conclusion
The maximum sum \( S_n \) occurs when \( n = 6 \) because all terms \( t_1, t_2, t_3, t_4, t_5, t_6 \) are positive, and \( t_7 \) is the first negative term.

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