The geometric mean of the first $n$ terms of the sequence $a, ar, ar^2, \dots$ is:

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

If the sum of infinite terms of a $G.P.$ is $3$ and the sum of the squares of its terms is $3$,then its first term and common ratio are:

If the $4^{\text{th}}$,$10^{\text{th}}$,and $16^{\text{th}}$ terms of a $G.P.$ are $x, y$,and $z$ respectively,prove that $x, y, z$ are in $G.P.$

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

Let $a_{1}, a_{2}, a_{3}, \dots$ be a $G$.$P$. of increasing positive terms such that $a_{2} \cdot a_{3} \cdot a_{4} = 64$ and $a_{1} + a_{3} + a_{5} = \frac{813}{7}$. Then $a_{3} + a_{5} + a_{7}$ is equal to: