The 4th term of an AP is 11. The sum of the 5th and 7th terms of this AP is 34.Find its common difference.

Explanation

To solve the problem step by step, we will use the formulas related to the terms of an Arithmetic Progression (AP).

### Step 1: Understand the given information
We know:
- The 4th term of the AP (T4) is 11.
- The sum of the 5th term (T5) and the 7th term (T7) is 34.

### Step 2: Use the formula for the nth term of an AP
The nth term of an AP can be expressed as:
\[ T_n = A + (n-1)D \]
where:
- \( A \) is the first term,
- \( D \) is the common difference,
- \( n \) is the term number.

### Step 3: Write the equations for T4, T5, and T7
1. For the 4th term (T4):
\[ T_4 = A + (4-1)D = A + 3D \]
Given that \( T_4 = 11 \), we have:
\[ A + 3D = 11 \quad \text{(Equation 1)} \]

2. For the 5th term (T5):
\[ T_5 = A + (5-1)D = A + 4D \]

3. For the 7th term (T7):
\[ T_7 = A + (7-1)D = A + 6D \]

### Step 4: Write the equation for the sum of T5 and T7
The sum of the 5th and 7th terms is given as:
\[ T_5 + T_7 = 34 \]
Substituting the expressions for T5 and T7:
\[ (A + 4D) + (A + 6D) = 34 \]
This simplifies to:
\[ 2A + 10D = 34 \quad \text{(Equation 2)} \]

### Step 5: Solve the equations
Now we have two equations:
1. \( A + 3D = 11 \) (Equation 1)
2. \( 2A + 10D = 34 \) (Equation 2)

From Equation 1, we can express \( A \) in terms of \( D \):
\[ A = 11 - 3D \]

Now, substitute this expression for \( A \) into Equation 2:
\[ 2(11 - 3D) + 10D = 34 \]
Expanding this gives:
\[ 22 - 6D + 10D = 34 \]
Combining like terms:
\[ 22 + 4D = 34 \]
Subtracting 22 from both sides:
\[ 4D = 12 \]
Dividing by 4:
\[ D = 3 \]

### Final Answer
The common difference \( D \) is 3.

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