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

$\text{Given: A.P. of natural numbers}$

$\dfrac{S_7}{S_{11}} = \dfrac{6}{11}$

$130 < a_7 < 140$

$\text{To find: the common difference } d$

$a_1 = a, \quad d = \text{common difference}$

$a_n = a + (n-1)d$

$d \in \mathbb{Z}$

$S_n = \dfrac{n}{2}\,[2a + (n-1)d]$

$S_7 = \dfrac{7}{2}(2a + 6d) = 7(a + 3d)$

$S_{11} = \dfrac{11}{2}(2a + 10d) = 11(a + 5d)$

$\dfrac{S_7}{S_{11}} = \dfrac{7(a+3d)}{11(a+5d)} = \dfrac{6}{11}$

$7(a+3d) = 6(a+5d)$

$7a + 21d = 6a + 30d$

$a = 9d$

$a_7 = a + 6d = 9d + 6d = 15d$

$130 < 15d < 140$

$\dfrac{130}{15} < d < \dfrac{140}{15}$

$8.67 < d < 9.33$

$d = 9 \quad (\text{integer})$

$a = 9 \times 9 = 81, \quad a_7 = 81 + 54 = 135$

$\boxed{d = 9}$