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

The sum of a $G.P.$ with common ratio $r = 3$ is $364$,and the last term is $243$. Find the number of terms $n$.

Let $n (> 1)$ be a positive integer. The largest integer $m$ such that $(n^m + 1)$ divides $(1 + n + n^2 + \dots + n^{127})$ is:

If $a, b, c, d$ and $p$ are distinct real numbers such that $(a^2 + b^2 + c^2)p^2 - 2p(ab + bc + cd) + (b^2 + c^2 + d^2) \le 0$,then

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 $a_1, a_2, a_3, \ldots$ be a $G.P.$ of increasing positive numbers. Let the sum of its $6^{\text{th}}$ and $8^{\text{th}}$ terms be $2$ and the product of its $3^{\text{rd}}$ and $5^{\text{th}}$ terms be $\frac{1}{9}$. Then $6(a_2 + a_4)(a_4 + a_6)$ is equal to