The $4^{\text {tht }}$ term of $GP$ is $500$ and its common ratio is $\frac{1}{m}, m \in N$. Let $S_n$ denote the sum of the first $n$ terms of this GP. If $S_6 > S_5+1$ and $S_7 < S_6+\frac{1}{2}$, then the number of possible values of $m$ is $……….$

Let ${A_n} = \left( {\frac{3}{4}} \right) – {\left( {\frac{3}{4}} \right)^2} + {\left( {\frac{3}{4}} \right)^3} – ….. + {\left( { – 1} \right)^{n – 1}}{\left( {\frac{3}{4}} \right)^n}$ and $B_n \,= 1 – A_n$ . Then, the least odd natural number $p$ , so that ${B_n} > {A_n}$, for all $n \geq p$ is

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 $c$

If the sum of the second, third and fourth terms of a positive term $G.P.$ is $3$ and the sum of its sixth, seventh and eighth terms is $243,$ then the sum of the first $50$ terms of this $G.P.$ is

Let $a_1, a_2, a_3, \ldots$. be a $GP$ of increasing positive numbers. If the product of fourth and sixth terms is $9$ and the sum of fifth and seventh terms is $24 ,$ then $a_1 a_9+a_2 a_4 a_9+a_5+a_7$ is equal to $………$.