Suppose that all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is 6: 11 and the seventh term lies in between 130 and 140, then the common difference of this A.P. is

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We know that:
- The ratio of the sum of the first 7 terms (S7) to the sum of the first 11 terms (S11) is 6:11.
- The 7th term lies between 130 and 140.

### Step 2: Write the formulas for the sums
The sum of the first n terms of an arithmetic progression (A.P.) can be expressed as:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
where \( a \) is the first term and \( d \) is the common difference.

For the first 7 terms:
\[ S_7 = \frac{7}{2} \times (2a + 6d) = \frac{7}{2} \times (2a + 6d) \]

For the first 11 terms:
\[ S_{11} = \frac{11}{2} \times (2a + 10d) = \frac{11}{2} \times (2a + 10d) \]

### Step 3: Set up the equation based on the given ratio
According to the problem, we have:
\[ \frac{S_7}{S_{11}} = \frac{6}{11} \]

Substituting the expressions for \( S_7 \) and \( S_{11} \):
\[ \frac{\frac{7}{2} \times (2a + 6d)}{\frac{11}{2} \times (2a + 10d)} = \frac{6}{11} \]

### Step 4: Simplify the equation
Cancelling out \( \frac{1}{2} \) from both sides, we get:
\[ \frac{7(2a + 6d)}{11(2a + 10d)} = \frac{6}{11} \]

Cross-multiplying gives:
\[ 7(2a + 6d) = 6(2a + 10d) \]

### Step 5: Expand and rearrange the equation
Expanding both sides:
\[ 14a + 42d = 12a + 60d \]

Rearranging gives:
\[ 14a - 12a = 60d - 42d \]
\[ 2a = 18d \]
\[ a = 9d \]

### Step 6: Find the 7th term
The 7th term of the A.P. is given by:
\[ T_7 = a + 6d \]
Substituting \( a = 9d \):
\[ T_7 = 9d + 6d = 15d \]

### Step 7: Set up the inequality for the 7th term
We know that:
\[ 130 < T_7 < 140 \]
Substituting \( T_7 = 15d \):
\[ 130 < 15d < 140 \]

### Step 8: Solve for \( d \)
Dividing the entire inequality by 15:
\[ \frac{130}{15} < d < \frac{140}{15} \]
Calculating the values:
\[ 8.67 < d < 9.33 \]

Since \( d \) must be a natural number, the only possible value for \( d \) is:
\[ d = 9 \]

### Conclusion
The common difference \( d \) of the A.P. is \( 9 \).

---