The third term of an arithmetical progression is 7, and the seventh term is 2 more than 3 times the third term. Find the first term, the common difference and the sum of the first 20 terms.

Answer

To solve the problem step by step, we will use the properties of an arithmetic progression (AP).

### Step 1: Define the terms of the AP
Let:
- \( a \) = first term of the AP
- \( d \) = common difference

The \( n \)-th term of an arithmetic progression can be expressed as:
\[ T_n = a + (n - 1)d \]

### Step 2: Write the equations for the given terms
1. The third term \( T_3 \) is given as 7:
\[ T_3 = a + (3 - 1)d = a + 2d = 7 \quad \text{(Equation 1)} \]

2. The seventh term \( T_7 \) is given as 2 more than 3 times the third term:
\[ T_7 = a + (7 - 1)d = a + 6d \]
According to the problem, \( T_7 = 2 + 3 \times T_3 \):
\[ T_7 = 2 + 3 \times 7 = 2 + 21 = 23 \quad \text{(Equation 2)} \]
Thus, we have:
\[ a + 6d = 23 \quad \text{(Equation 2)} \]

### Step 3: Solve the equations
Now we have two equations:
1. \( a + 2d = 7 \) (Equation 1)
2. \( a + 6d = 23 \) (Equation 2)

We can subtract Equation 1 from Equation 2 to eliminate \( a \):
\[ (a + 6d) - (a + 2d) = 23 - 7 \]
This simplifies to:
\[ 4d = 16 \]
Dividing both sides by 4 gives:
\[ d = 4 \]

### Step 4: Find the first term \( a \)
Now that we have \( d \), we can substitute it back into Equation 1 to find \( a \):
\[ a + 2(4) = 7 \]
\[ a + 8 = 7 \]
\[ a = 7 - 8 = -1 \]

### Step 5: Calculate the sum of the first 20 terms
The sum of the first \( n \) terms of an arithmetic progression is given by:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]
For \( n = 20 \):
\[ S_{20} = \frac{20}{2} \times (2(-1) + (20 - 1)(4)) \]
Calculating this step by step:
\[ S_{20} = 10 \times (2(-1) + 19 \times 4) \]
\[ = 10 \times (-2 + 76) \]
\[ = 10 \times 74 \]
\[ = 740 \]

### Final Results
- First term \( a = -1 \)
- Common difference \( d = 4 \)
- Sum of the first 20 terms \( S_{20} = 740 \)