If the sum of the first 2n terms of the AP 2, 5, 8 ....is equal to the sum of first n terms of the AP 57, 59, 61, ..., then what is the value of n?

Options

A. 7
B. 9
C. 11
D. 13

Answer

To solve the problem, we need to find the value of \( n \) such that the sum of the first \( 2n \) terms of the first arithmetic progression (AP) is equal to the sum of the first \( n \) terms of the second AP.

### Step-by-Step Solution:

1. **Identify the first AP**: The first AP is given as \( 2, 5, 8, \ldots \).
- First term \( a_1 = 2 \)
- Common difference \( d_1 = 5 - 2 = 3 \)

2. **Find the sum of the first \( 2n \) terms of the first AP**:
- The formula for the sum of the first \( n \) terms of an AP is:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]
- For the first AP, substituting \( n = 2n \):
\[ S_{2n} = \frac{2n}{2} \times \left(2 \cdot 2 + (2n - 1) \cdot 3\right) \]
\[ = n \times \left(4 + (2n - 1) \cdot 3\right) \]
\[ = n \times \left(4 + 6n - 3\right) \]
\[ = n \times (6n + 1) \]

3. **Identify the second AP**: The second AP is given as \( 57, 59, 61, \ldots \).
- First term \( a_2 = 57 \)
- Common difference \( d_2 = 59 - 57 = 2 \)

4. **Find the sum of the first \( n \) terms of the second AP**:
- For the second AP, substituting \( n \):
\[ S_n = \frac{n}{2} \times (2 \cdot 57 + (n - 1) \cdot 2) \]
\[ = \frac{n}{2} \times (114 + 2n - 2) \]
\[ = \frac{n}{2} \times (2n + 112) \]
\[ = n \times (n + 56) \]

5. **Set the two sums equal to each other**:
\[ n \times (6n + 1) = n \times (n + 56) \]

6. **Assuming \( n \neq 0 \), we can divide both sides by \( n \)**:
\[ 6n + 1 = n + 56 \]

7. **Rearranging the equation**:
\[ 6n - n = 56 - 1 \]
\[ 5n = 55 \]
\[ n = \frac{55}{5} = 11 \]

### Final Answer:
The value of \( n \) is \( 11 \).