The 17th term of an A.P. exceeds its 10th term by 7. Find the common difference.

Answer

To solve the problem, we need to find the common difference of an arithmetic progression (A.P.) given that the 17th term exceeds the 10th term by 7.

### Step-by-Step Solution:

1. **Understand the formula for the nth term of an A.P.**:
The nth term of an A.P. can be expressed as:
\[ A_n = A + (n - 1) \cdot d \]
where \( A \) is the first term, \( d \) is the common difference, and \( n \) is the term number.

2. **Write the expressions for the 17th and 10th terms**:
- The 17th term \( A_{17} \) is:
\[ A_{17} = A + (17 - 1) \cdot d = A + 16d \]
- The 10th term \( A_{10} \) is:
\[ A_{10} = A + (10 - 1) \cdot d = A + 9d \]

3. **Set up the equation based on the given information**:
According to the problem, the 17th term exceeds the 10th term by 7:
\[ A_{17} = A_{10} + 7 \]
Substituting the expressions for \( A_{17} \) and \( A_{10} \):
\[ A + 16d = (A + 9d) + 7 \]

4. **Simplify the equation**:
Rearranging the equation gives:
\[ A + 16d = A + 9d + 7 \]
We can subtract \( A \) from both sides:
\[ 16d = 9d + 7 \]

5. **Isolate the common difference \( d \)**:
Subtract \( 9d \) from both sides:
\[ 16d - 9d = 7 \]
This simplifies to:
\[ 7d = 7 \]

6. **Solve for \( d \)**:
Dividing both sides by 7 gives:
\[ d = 1 \]

### Final Answer:
The common difference \( d \) is \( 1 \).