In an A.P., `T_(3) = 7 and T_(7) = 2 + 3 T_(3)`, then sum of its first 20 terms is :

Options

A. 74
B. 228
C. 740
D. 1090

Answer

To solve the problem, we need to find the sum of the first 20 terms of an arithmetic progression (A.P.) given that \( T_3 = 7 \) and \( T_7 = 2 + 3T_3 \).

### Step-by-Step Solution:

1. **Write the general formula for the nth term of an A.P.:**
\[ T_n = a + (n-1)d \]
where \( a \) is the first term and \( d \) is the common difference.

2. **Set up the equations using the given terms:**
- For \( T_3 = 7 \):
\[ T_3 = a + 2d = 7 \quad \text{(Equation 1)} \]
- For \( T_7 = 2 + 3T_3 \):
\[ T_7 = a + 6d = 2 + 3 \times 7 = 23 \quad \text{(Equation 2)} \]

3. **Now we have two equations:**
- Equation 1: \( a + 2d = 7 \)
- Equation 2: \( a + 6d = 23 \)

4. **Subtract Equation 1 from Equation 2 to eliminate \( a \):**
\[ (a + 6d) - (a + 2d) = 23 - 7 \]
Simplifying this gives:
\[ 4d = 16 \]
Therefore, we find:
\[ d = 4 \]

5. **Substitute \( d \) back into Equation 1 to find \( a \):**
\[ a + 2(4) = 7 \]
This simplifies to:
\[ a + 8 = 7 \implies a = 7 - 8 = -1 \]

6. **Now we have the first term \( a = -1 \) and the common difference \( d = 4 \).**

7. **To find the sum of the first 20 terms \( S_{20} \):**
The formula for the sum of the first \( n \) terms of an A.P. is:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
For \( n = 20 \):
\[ S_{20} = \frac{20}{2} \times (2(-1) + (20-1)(4)) \]
Simplifying this gives:
\[ S_{20} = 10 \times (-2 + 19 \times 4) \]
Calculate \( 19 \times 4 = 76 \):
\[ S_{20} = 10 \times (-2 + 76) = 10 \times 74 = 740 \]

### Final Answer:
The sum of the first 20 terms is \( \boxed{740} \).