The fourth term of an A.P. is 11 and the eighth term exceeds twice the fourth term by 5. Find the A.P. and the sum of first 50 terms.

Answer

To solve the problem step by step, we will use the properties of an Arithmetic Progression (A.P.).

### Step 1: Understand the terms of A.P.
The n-th term of an A.P. can be expressed as:
\[ T_n = a + (n-1)d \]
where \( a \) is the first term and \( d \) is the common difference.

### Step 2: Set up the equations based on the given information.
We are given:
- The fourth term \( T_4 = 11 \)
- The eighth term \( T_8 \) exceeds twice the fourth term by 5.

From the first piece of information:
\[ T_4 = a + 3d = 11 \quad (1) \]

From the second piece of information:
\[ T_8 = a + 7d \]
Since \( T_8 \) exceeds twice the fourth term by 5, we can write:
\[ T_8 = 2 \times T_4 + 5 \]
Substituting \( T_4 = 11 \):
\[ T_8 = 2 \times 11 + 5 = 22 + 5 = 27 \quad (2) \]

### Step 3: Set up the second equation.
From equation (2):
\[ a + 7d = 27 \quad (3) \]

### Step 4: Solve the system of equations.
Now we have two equations:
1. \( a + 3d = 11 \) (equation 1)
2. \( a + 7d = 27 \) (equation 3)

Subtract equation (1) from equation (3):
\[ (a + 7d) - (a + 3d) = 27 - 11 \]
This simplifies to:
\[ 4d = 16 \]
Thus, we find:
\[ d = 4 \]

### Step 5: Substitute \( d \) back to find \( a \).
Now substitute \( d = 4 \) back into equation (1):
\[ a + 3(4) = 11 \]
\[ a + 12 = 11 \]
\[ a = 11 - 12 = -1 \]

### Step 6: Write the A.P.
Now we have \( a = -1 \) and \( d = 4 \). The first few terms of the A.P. are:
- \( T_1 = a = -1 \)
- \( T_2 = a + d = -1 + 4 = 3 \)
- \( T_3 = a + 2d = -1 + 2 \times 4 = 7 \)
- \( T_4 = 11 \)
- \( T_5 = a + 4d = -1 + 4 \times 4 = 15 \)
- And so on...

The A.P. is: \(-1, 3, 7, 11, 15, \ldots\)

### Step 7: Find the sum of the first 50 terms.
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 \( n = 50 \):
\[ S_{50} = \frac{50}{2} \times (2(-1) + (50-1) \times 4) \]
\[ = 25 \times (-2 + 49 \times 4) \]
Calculating \( 49 \times 4 = 196 \):
\[ S_{50} = 25 \times (-2 + 196) = 25 \times 194 \]
Calculating \( 25 \times 194 = 4850 \).

### Final Answer:
The A.P. is \(-1, 3, 7, 11, 15, \ldots\) and the sum of the first 50 terms is \( 4850 \).

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