Let $a_1, \frac{a_2}{2}, \frac{a_3}{2^2}, \ldots, \frac{a_{10}}{2^9}$ be a $G$.$P$. with common ratio $r = \frac{1}{\sqrt{2}}$. If $a_1 + a_2 + \ldots + a_{10} = 62$, then $a_1$ is equal to:

Insert two numbers between $3$ and $81$ so that the resulting sequence is a $G.P.$

If $x, y, z$ are in geometric progression and $a^x = b^y = c^z$,then . . . . . .

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 $b_1, b_2, b_3, b_4$ are neither in $A.P.$ nor in $G.P.$
$STATEMENT-2$ : The numbers $b_1, b_2, b_3, b_4$ are in $H.P.$

Six positive numbers are in $GP$, such that their product is $1000$. If the fourth term is $1$, then the last term is

The arithmetic mean of the sequence $1, 2, 4, 8, 16, \ldots, 2^n$ is: