If first, second and last terms of an A.P. are a, b and 2a respectively, then find the sum of first n terms.

Options

A. `(3ab)/(2(a+b))`
B. `(3ab)/(a+b)`
C. `(3ab)/(a-b)`
D. `(3ab)/(2(b-a))`

Answer

To find the sum of the first \( n \) terms of an arithmetic progression (A.P.) where the first term is \( a \), the second term is \( b \), and the last term is \( 2a \), we will follow these steps:

### Step 1: Identify the terms of the A.P.
The first term \( A_1 = a \), the second term \( A_2 = b \), and the last term \( A_n = 2a \).

### Step 2: Find the common difference \( d \).
The common difference \( d \) can be calculated as:
\[ d = A_2 - A_1 = b - a \]

### Step 3: Use the formula for the \( n \)-th term of an A.P.
The formula for the \( n \)-th term of an A.P. is given by:
\[ A_n = A_1 + (n-1)d \]
Substituting the known values:
\[ 2a = a + (n-1)(b - a) \]

### Step 4: Solve for \( n \).
Rearranging the equation:
\[ 2a - a = (n-1)(b - a) \]
\[ a = (n-1)(b - a) \]
Now, divide both sides by \( (b - a) \):
\[ \frac{a}{b - a} = n - 1 \]
Thus,
\[ n = \frac{a}{b - a} + 1 \]

### Step 5: Use the sum formula for the first \( n \) terms of an A.P.
The sum \( S_n \) of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} (A_1 + A_n) \]
Substituting \( A_1 = a \) and \( A_n = 2a \):
\[ S_n = \frac{n}{2} (a + 2a) = \frac{n}{2} (3a) = \frac{3an}{2} \]

### Step 6: Substitute \( n \) into the sum formula.
Now substituting the value of \( n \):
\[ S_n = \frac{3a}{2} \left( \frac{a}{b - a} + 1 \right) \]
This simplifies to:
\[ S_n = \frac{3a}{2} \left( \frac{a + (b - a)}{b - a} \right) = \frac{3a}{2} \left( \frac{b}{b - a} \right) \]

### Final Result:
Thus, the sum of the first \( n \) terms is:
\[ S_n = \frac{3ab}{2(b - a)} \]