If the `10^(th)` term of an A.P. is `1/5` and its `20^(th)` term is `1/3` , then the sum of its first 150 terms is:

Options

A. `169`
B. `161`
C. `164`
D. `167`

Explanation

To solve the problem, we need to find the sum of the first 150 terms of an arithmetic progression (A.P.) given the 10th term and the 20th term.

### Step-by-step Solution:

1. **Identify the terms given:**
- The 10th term \( a_{10} = \frac{1}{5} \)
- The 20th term \( a_{20} = \frac{1}{3} \)

2. **Use the formula for the nth term of an A.P.:**
The nth term of an A.P. is given by:
\[ a_n = a + (n - 1)d \]
where \( a \) is the first term and \( d \) is the common difference.

3. **Set up equations for the given terms:**
For the 10th term:
\[ a + 9d = \frac{1}{5} \quad \text{(1)} \]
For the 20th term:
\[ a + 19d = \frac{1}{3} \quad \text{(2)} \]

4. **Subtract equation (1) from equation (2):**
\[ (a + 19d) - (a + 9d) = \frac{1}{3} - \frac{1}{5} \]
Simplifying this gives:
\[ 10d = \frac{1}{3} - \frac{1}{5} \]
To subtract the fractions, find a common denominator (15):
\[ \frac{1}{3} = \frac{5}{15}, \quad \frac{1}{5} = \frac{3}{15} \]
Therefore:
\[ 10d = \frac{5}{15} - \frac{3}{15} = \frac{2}{15} \]
Thus:
\[ d = \frac{2}{150} = \frac{1}{75} \]

5. **Substitute \( d \) back into one of the equations to find \( a \):**
Using equation (1):
\[ a + 9d = \frac{1}{5} \]
Substitute \( d = \frac{1}{75} \):
\[ a + 9 \times \frac{1}{75} = \frac{1}{5} \]
This simplifies to:
\[ a + \frac{9}{75} = \frac{1}{5} \]
Convert \( \frac{1}{5} \) to have a denominator of 75:
\[ \frac{1}{5} = \frac{15}{75} \]
Therefore:
\[ a + \frac{9}{75} = \frac{15}{75} \]
Thus:
\[ a = \frac{15}{75} - \frac{9}{75} = \frac{6}{75} = \frac{2}{25} \]

6. **Calculate the sum of the first 150 terms:**
The sum \( S_n \) of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
For \( n = 150 \):
\[ S_{150} = \frac{150}{2} \times \left(2 \times \frac{2}{25} + (150 - 1) \times \frac{1}{75}\right) \]
Simplifying:
\[ S_{150} = 75 \times \left(\frac{4}{25} + 149 \times \frac{1}{75}\right) \]
Convert \( \frac{4}{25} \) to have a denominator of 75:
\[ \frac{4}{25} = \frac{12}{75} \]
Therefore:
\[ S_{150} = 75 \times \left(\frac{12}{75} + \frac{149}{75}\right) = 75 \times \frac{161}{75} \]
Thus:
\[ S_{150} = 161 \]

### Final Result:
The sum of the first 150 terms of the A.P. is \( \boxed{161} \).