$${{{S_7}} \over {{S_{11}}}} = {6 \over {11}}$$ ...... (1)
$$130 \le {t_7} \le 140$$ ........ (2)
$$ \Rightarrow {{{7 \over 2}[2a + 6d]} \over {{{11} \over 2}[2a + 10d]}} = {6 \over {11}}$$
$$ \Rightarrow {{a + 3d} \over {a + 5d}} = {6 \over 7}$$ ....... (3)
$$ \Rightarrow {{{t_4}} \over {{4_6}}} = {6 \over 7}$$
Let $${t_4} = 6k$$, $${t_6} = 7k$$;
$$2d = k \Rightarrow d = k/2$$ and $$a + 3d = 6k$$
$$ \Rightarrow a = 6k - 3k/2 = 9k/2$$
Hence, $$130 \le {t_7} \le 140$$.
$$ \Rightarrow 130 \le {{9k} \over 2} + 3k \le 140$$
$$ \Rightarrow 130 \le {{15k} \over 2} \le 140$$
$$ \Rightarrow {{52} \over 3} \le k \in {{56} \over 3}$$
Since, $$k \in N \Rightarrow k = 18$$.
$$ \Rightarrow d = {k \over 2} = {{18} \over 2} = 9$$