Find the sum of the following A. P. s:
` (1)/(15) ,(1)/(12), (1)/(10),…,` to 11 terms

Answer

To find the sum of the given arithmetic progression (A.P.): \( \frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \ldots \) for the first 11 terms, we will follow these steps:

### Step 1: Identify the first term (a) and the common difference (d)
The first term \( a \) of the A.P. is:
\[ a = \frac{1}{15} \]

Next, we need to find the common difference \( d \). The common difference is calculated by subtracting the first term from the second term:
\[ d = \frac{1}{12} - \frac{1}{15} \]

To perform this subtraction, we need a common denominator. The least common multiple of 12 and 15 is 60. Thus, we convert both fractions:
\[ \frac{1}{12} = \frac{5}{60}, \quad \frac{1}{15} = \frac{4}{60} \]
Now, we can find \( d \):
\[ d = \frac{5}{60} - \frac{4}{60} = \frac{1}{60} \]

### Step 2: Use the formula for the sum of the first n terms of an A.P.
The formula for the sum \( S_n \) of the first \( n \) terms of an A.P. is:
\[ S_n = \frac{n}{2} \left( 2a + (n-1)d \right) \]
Here, \( n = 11 \), \( a = \frac{1}{15} \), and \( d = \frac{1}{60} \).

### Step 3: Substitute the values into the formula
Substituting the values into the formula:
\[ S_{11} = \frac{11}{2} \left( 2 \times \frac{1}{15} + (11-1) \times \frac{1}{60} \right) \]

Calculating \( 2a \):
\[ 2a = 2 \times \frac{1}{15} = \frac{2}{15} \]

Calculating \( (n-1)d \):
\[ (n-1)d = 10 \times \frac{1}{60} = \frac{10}{60} = \frac{1}{6} \]

### Step 4: Combine the terms inside the parentheses
Now we need to add \( \frac{2}{15} \) and \( \frac{1}{6} \). The least common multiple of 15 and 6 is 30:
\[ \frac{2}{15} = \frac{4}{30}, \quad \frac{1}{6} = \frac{5}{30} \]
Thus,
\[ \frac{2}{15} + \frac{1}{6} = \frac{4}{30} + \frac{5}{30} = \frac{9}{30} = \frac{3}{10} \]

### Step 5: Substitute back into the sum formula
Now substituting back into the sum formula:
\[ S_{11} = \frac{11}{2} \left( \frac{3}{10} \right) = \frac{11 \times 3}{2 \times 10} = \frac{33}{20} \]

### Final Answer
The sum of the first 11 terms of the A.P. is:
\[ \boxed{\frac{33}{20}} \]