The sum of first `2n` terms of an `AP` is `alpha`. and the sum of next `n` terms is `beta,` its common difference is

Options

A. `(alpha-2beta)/(3n^(2))`
B. `(2beta-alpha)/(3n^(2))`
C. `(alpha-2beta)/(3n)`
D. `(2beta-alpha)/(3n)`

Answer

To find the common difference \( d \) of the arithmetic progression (AP) given the sum of the first \( 2n \) terms is \( \alpha \) and the sum of the next \( n \) terms is \( \beta \), we can follow these steps:

### Step 1: Sum of the first \( 2n \) terms
The formula for the sum of the first \( n \) terms of an AP is given by:

\[ S_n = \frac{n}{2} \left(2a + (n-1)d\right) \]

For the first \( 2n \) terms, we have:

\[ S_{2n} = \frac{2n}{2} \left(2a + (2n-1)d\right) = n \left(2a + (2n-1)d\right) \]

According to the problem, this sum is equal to \( \alpha \):

\[ n \left(2a + (2n-1)d\right) = \alpha \]

### Step 2: Sum of the next \( n \) terms
The sum of the next \( n \) terms (from \( 2n+1 \) to \( 3n \)) can be calculated as follows:

\[ S_{n} = \frac{n}{2} \left(2a + (2n)d + (n-1)d\right) = \frac{n}{2} \left(2a + (2n+n-1)d\right) = \frac{n}{2} \left(2a + (3n-1)d\right) \]

This sum is given as \( \beta \):

\[ \frac{n}{2} \left(2a + (3n-1)d\right) = \beta \]

### Step 3: Setting up the equations
Now we have two equations:

1. \( n(2a + (2n-1)d) = \alpha \)
2. \( \frac{n}{2}(2a + (3n-1)d) = \beta \)

### Step 4: Solving for \( a \)
From the first equation, we can express \( 2a \):

\[ 2a + (2n-1)d = \frac{\alpha}{n} \]

Thus,

\[ 2a = \frac{\alpha}{n} - (2n-1)d \tag{1} \]

### Step 5: Substitute \( a \) in the second equation
Now substitute \( 2a \) from equation (1) into the second equation:

\[ \frac{n}{2} \left(\frac{\alpha}{n} - (2n-1)d + (3n-1)d\right) = \beta \]

This simplifies to:

\[ \frac{n}{2} \left(\frac{\alpha}{n} + (n+2)d\right) = \beta \]

### Step 6: Multiply through by 2
Multiply both sides by 2 to eliminate the fraction:

\[ n \left(\frac{\alpha}{n} + (n+2)d\right) = 2\beta \]

This simplifies to:

\[ \alpha + n(n+2)d = 2\beta \]

### Step 7: Solve for \( d \)
Rearranging gives:

\[ n(n+2)d = 2\beta - \alpha \]

Thus,

\[ d = \frac{2\beta - \alpha}{n(n+2)} \]

### Final Answer
The common difference \( d \) is:

\[ d = \frac{2\beta - \alpha}{n(n+2)} \]