Find a $G.P.$ for which sum of the first two terms is $-4$ and the fifth term is $4$ times the third term.
Let $a$ be the first term and $r$ be the common ratio of the $G.P.$
According to the given conditions,
$A_{2}=-4=\frac{a\left(1-r^{2}\right)}{1-r}$ .......$(1)$
$a_{5}=4 \times a_{3}$
$\Rightarrow a r^{4}=4 a r^{2} \Rightarrow r^{2}=4$
$\therefore r=\pm 2$
From $(1),$ we obtain
$-4=\frac{a\left[1-(2)^{2}\right]}{1-2}$ for $r=2$
$\Rightarrow-4=\frac{a(1-4)}{-1}$
$\Rightarrow-4=a(3)$
$\Rightarrow a=\frac{-4}{3}$
Also, $-4=\frac{a\left[1-(-2)^{2}\right]}{1-(-2)}$ for $r=-2$
$\Rightarrow-4=\frac{a(1-4)}{1+2}$
$\Rightarrow-4=\frac{a(-3)}{3}$
$\Rightarrow a=4$
Thus, the required $G.P.$ is $\frac{-4}{3}, \frac{-8}{3}, \frac{-16}{3}, \ldots$ or $4,-8,-16,-32 \ldots$
Suppose four distinct positive numbers $a_1, a_2, a_3, a_4$ are in $G.P.$ Let $b_1=a_1, b_2=b_1+a_2, b_3=b_2+a_3$ and $b_4=b_3+a_4$.
$STATEMENT-1$ : The numbers $\mathrm{b}_1, \mathrm{~b}_2, \mathrm{~b}_3, \mathrm{~b}_4$ are neither in $A.P$. nor in $G.P.$ and
$STATEMENT-2$ : The numbers $\mathrm{b}_1, \mathrm{~b}_2, \mathrm{~b}_3, \mathrm{~b}_4$ are in $H.P.$
If ${s_n} = 1 + \frac{1}{2} + \frac{1}{{{2^2}}} + ........ + \frac{1}{{{2^{n - 1}}}}$ , then the least integral value of $n$ such that $2 - {s_n} < \frac{1}{{100}}$ is
The sum of infinite terms of a $G.P.$ is $x$ and on squaring the each term of it, the sum will be $y$, then the common ratio of this series 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 $c$
In a $G.P.,$ the $3^{rd}$ term is $24$ and the $6^{\text {th }}$ term is $192 .$ Find the $10^{\text {th }}$ term.