If in a $G.P.$ of $64$ terms, the sum of all the terms is $7$ times the sum of the odd terms of the $G.P,$ then the common ratio of the $G.P$. is equal to

For $0<\mathrm{c}<\mathrm{b}<\mathrm{a}$, let $(\mathrm{a}+\mathrm{b}-2 \mathrm{c}) \mathrm{x}^2+(\mathrm{b}+\mathrm{c}-2 \mathrm{a}) \mathrm{x}$ $+(c+a-2 b)=0$ and $\alpha \neq 1$ be one of its root. Then, among the two statements

$(I)$ If $\alpha \in(-1,0)$, then $\mathrm{b}$ cannot be the geometric mean of $\mathrm{a}$ and $\mathrm{c}$

$(II)$ If $\alpha \in(0,1)$, then $\mathrm{b}$ may be the geometric mean of $a$ and $c$

The sum of first $20$ terms of the sequence $0.7,0.77,0.777, . . . $ is 

$0.\mathop {423}\limits^{\,\,\,\,\, \bullet \, \bullet \,}  = $

If the sum of the $n$ terms of $G.P.$ is $S$ product is $P$ and sum of their inverse is $R$, than ${P^2}$ is equal to

Let $P(x)=1+x+x^2+x^3+x^4+x^5$. What is the remainder when $P\left(x^{12}\right)$ is divided by $P(x)$ ?