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 that the 10th term is \( \frac{1}{5} \) and the 20th term is \( \frac{1}{3} \).

### Step 1: Write the formulas for the terms of the A.P.
The nth term of an A.P. can be expressed as:
\[ T_n = a + (n-1)d \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 2: Set up equations for the given terms.
From the problem, we have:
- For the 10th term:
\[ T_{10} = a + 9d = \frac{1}{5} \quad \text{(1)} \]
- For the 20th term:
\[ T_{20} = a + 19d = \frac{1}{3} \quad \text{(2)} \]

### Step 3: Subtract the equations to find \( d \).
Subtract equation (1) from equation (2):
\[ T_{20} - T_{10} = (a + 19d) - (a + 9d) = 10d \]
Thus,
\[ \frac{1}{3} - \frac{1}{5} = 10d \]

### Step 4: Simplify the left-hand side.
To subtract the fractions, find a common denominator (which is 15):
\[ \frac{1}{3} = \frac{5}{15}, \quad \frac{1}{5} = \frac{3}{15} \]
So,
\[ \frac{5}{15} - \frac{3}{15} = \frac{2}{15} \]
Thus, we have:
\[ 10d = \frac{2}{15} \]

### Step 5: Solve for \( d \).
Dividing both sides by 10:
\[ d = \frac{2}{15 \times 10} = \frac{2}{150} = \frac{1}{75} \]

### Step 6: Substitute \( d \) back to find \( a \).
Using equation (1):
\[ a + 9d = \frac{1}{5} \]
Substituting \( d \):
\[ a + 9 \left(\frac{1}{75}\right) = \frac{1}{5} \]
This simplifies to:
\[ a + \frac{9}{75} = \frac{1}{5} \]
Converting \( \frac{1}{5} \) to a fraction with a denominator of 75:
\[ \frac{1}{5} = \frac{15}{75} \]
So we have:
\[ a + \frac{9}{75} = \frac{15}{75} \]
Subtracting \( \frac{9}{75} \) from both sides:
\[ a = \frac{15}{75} - \frac{9}{75} = \frac{6}{75} = \frac{2}{25} \]

### Step 7: Calculate the sum of the first 150 terms.
The sum of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]
For \( n = 150 \):
\[ S_{150} = \frac{150}{2} \left(2 \cdot \frac{2}{25} + (150 - 1) \cdot \frac{1}{75}\right) \]
Calculating:
\[ S_{150} = 75 \left(\frac{4}{25} + 149 \cdot \frac{1}{75}\right) \]
Calculating \( 149 \cdot \frac{1}{75} \):
\[ 149 \cdot \frac{1}{75} = \frac{149}{75} \]
Now, convert \( \frac{4}{25} \) to a fraction with a denominator of 75:
\[ \frac{4}{25} = \frac{12}{75} \]
Thus,
\[ S_{150} = 75 \left(\frac{12}{75} + \frac{149}{75}\right) = 75 \cdot \frac{161}{75} = 161 \]

### Final Answer:
The sum of the first 150 terms is \( 161 \).

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