If the first second and the last terms of an A.P are a,band 2a , respectively, then its sum is

Options

A. `(ab)/(2(b-a))`
B. `(ab)/(b-a)`
C. `(3ab)/(2(b-a))`
D. None of these

Answer

To solve the problem, we need to find the sum of an arithmetic progression (A.P.) where the first term \( T_1 = a \), the second term \( T_2 = b \), and the last term \( T_n = 2a \).

### Step-by-step Solution:

1. **Identify the Terms:**
- First term \( T_1 = a \)
- Second term \( T_2 = b \)
- Last term \( T_n = 2a \)

2. **Find the Common Difference:**
The common difference \( d \) can be calculated as:
\[ d = T_2 - T_1 = b - a \]

3. **Use the Formula for the Last Term:**
The last term of an A.P. can be expressed as:
\[ T_n = T_1 + (n - 1)d \]
Substituting the known values:
\[ 2a = a + (n - 1)(b - a) \]

4. **Rearranging the Equation:**
Rearranging the equation gives:
\[ 2a - a = (n - 1)(b - a) \]
Simplifying this, we have:
\[ a = (n - 1)(b - a) \]

5. **Solving for \( n \):**
Rearranging further:
\[ n - 1 = \frac{a}{b - a} \]
Thus,
\[ n = \frac{a}{b - a} + 1 \]
This can also be written as:
\[ n = \frac{a + b - a}{b - a} = \frac{b}{b - a} \]

6. **Finding the Sum of the A.P.:**
The sum \( S_n \) of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} (T_1 + T_n) \]
Substituting the values we have:
\[ S_n = \frac{n}{2} (a + 2a) = \frac{n}{2} \cdot 3a = \frac{3a}{2} n \]

7. **Substituting for \( n \):**
Now substituting the value of \( n \):
\[ S_n = \frac{3a}{2} \cdot \frac{b}{b - a} \]

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