The sum of eleven terms of an A.P. whose middle terms is 30 is

Options

A. 320
B. 330
C. 340
D. 350

Answer

To find the sum of eleven terms of an arithmetic progression (A.P.) whose middle term is 30, we can follow these steps:

### Step 1: Identify the middle term
In an A.P. with an odd number of terms, the middle term is the term at position \( n/2 + 1 \). For 11 terms, the middle term is the 6th term.

\[ T_6 = a + (6 - 1)d = a + 5d \]

Given that the middle term \( T_6 = 30 \), we have:

\[ a + 5d = 30 \quad \text{(Equation 1)} \]

### Step 2: Write the formula for the sum of the first n 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 \( n = 11 \):

\[ S_{11} = \frac{11}{2} \times (2a + 10d) \]

### Step 3: Simplify the sum formula
We can rewrite the sum formula:

\[ S_{11} = \frac{11}{2} \times (2a + 10d) = \frac{11}{2} \times 2(a + 5d) = 11(a + 5d) \]

### Step 4: Substitute the value of \( a + 5d \)
From Equation 1, we know that \( a + 5d = 30 \). Therefore, we can substitute this value into the sum formula:

\[ S_{11} = 11 \times 30 \]

### Step 5: Calculate the sum
Now, we can calculate the sum:

\[ S_{11} = 330 \]

Thus, the sum of the eleven terms of the A.P. is **330**.