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, $$k \in N \Rightarrow k = 18$$.

$$ \Rightarrow d = {k \over 2} = {{18} \over 2} = 9$$