Given expression is
$$E = (1 + x)(1 + {x^2})(1 + {x^3})......(1 + {x^{100}})$$
Coefficient of x9 in E
= coefficient of x9 in $$(1 + x)(1 + {x^2})(1 + {x^3})......(1 + {x^9})$$
$$\Rightarrow$$ Terms containing x9
$$ = (1\,.\,{x^9} + {x^1}\,.\,{x^8} + {x^2}\,.\,{x^7} + {x^3}\,.\,{x^6} + {x^4}\,.\,{x^5} + {x^1}\,.\,{x^2}\,.\,{x^6} + {x^1}\,.\,{x^3}\,.\,{x^5} + {x^1}\,.\,{x^4}\,.\,{x^5})$$
$$\Rightarrow$$ Term containing x9 is 8x9 in E $$\Rightarrow$$ Coefficient of x9 = 8.
$${{{S_7}} \over {{S_{11}}}} = {6 \over {11}}$$ ...... (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, $$k \in N \Rightarrow k = 18$$.
$$ \Rightarrow d = {k \over 2} = {{18} \over 2} = 9$$