The two geometric means between $1$ and $64$ are ........

If the third term of a $G.P.$ is $4$,then the product of its first $5$ terms is:

How many terms of the $G.P.$ $3, 3^{2}, 3^{3}, \dots$ are needed to give the sum $120$?

The $G.M.$ of the numbers $3, 3^2, 3^3, ..., 3^n$ is

In an increasing geometric series,the sum of the second and the sixth term is $\frac{25}{2}$ and the product of the third and fifth term is $25$. Then,the sum of the $4^{\text{th}}$,$6^{\text{th}}$,and $8^{\text{th}}$ terms is equal to

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: