Common Difference of Arithmetic Progression of Natural Numbers

Let $a$ be the first term and $d$ be the common difference of the given Arithmetic Progression (A.P.). Since every term of the A.P. is a natural number, both $a$ and $d$ must be positive integers ($a, d \in \mathbb{N}$).

The sum of the first $n$ terms of an A.P. is given by the formula: $S_n = \frac{n}{2} [2a + (n - 1)d]$

Thus, the sum of the first 7 terms ($S_7$) and the sum of the first 11 terms ($S_{11}$) are: $S_7 = \frac{7}{2}(2a + 6d) = 7(a + 3d)$ $S_{11} = \frac{11}{2}(2a + 10d) = 11(a + 5d)$

We are given that the ratio of $S_7$ to $S_{11}$ is $6 : 11$: $\frac{S_7}{S_{11}} = \frac{6}{11}$

Substituting $S_7$ and $S_{11}$ into the ratio: $\frac{7(a + 3d)}{11(a + 5d)} = \frac{6}{11}$

Canceling $11$ from both denominators gives: $7(a + 3d) = 6(a + 5d)$

Expanding and simplifying the equation: $7a + 21d = 6a + 30d$ $a = 9d$

Next, the $7^{\text{th}}$ term of the A.P. is given by: $T_7 = a + 6d$

Substituting $a = 9d$ into the expression for $T_7$: $T_7 = 9d + 6d = 15d$

We are given that the $7^{\text{th}}$ term lies strictly between $130$ and $140$: $130 < T_7 < 140$ $130 < 15d < 140$

Dividing the entire inequality by $15$: $\frac{130}{15} < d < \frac{140}{15}$ $\frac{26}{3} < d < \frac{28}{3}$ $8.67 < d < 9.33$

Since $d$ must be a natural number, the only integer value in this open interval is: $d = 9$

(This yields $a = 9 \times 9 = 81$, which makes $T_7 = 135 \in (130, 140)$ and ensures all terms are natural numbers).

Therefore, the common difference of the A.P. is $9$.