The sum of first three terms of a $G.P.$ is $\frac{13}{12}$ and their product is $-1$ Find the common ratio and the terms.

If in a geometric progression $\left\{ {{a_n}} \right\},\;{a_1} = 3,\;{a_n} = 96$ and ${S_n} = 189$ then the value of $n$ is

Let $\alpha$ and $\beta$ be the roots of $x^{2}-3 x+p=0$ and $\gamma$ and $\delta$ be the roots of $x^{2}-6 x+q=0 .$ If $\alpha$ $\beta, \gamma, \delta$ form a geometric progression. Then ratio $(2 q+p):(2 q-p)$ is 

If three successive terms of a$G.P.$ with common ratio $r(r>1)$ are the lengths of the sides of a triangle and $[\mathrm{r}]$ denotes the greatest integer less than or equal to $r$, then $3[r]+[-r]$ is equal to :

If $a,b,c$ are in $A.P.$, then ${2^{ax + 1}},{2^{bx + 1}},\,{2^{cx + 1}},x \ne 0$ are in