In a geometric progression with $n$ terms and a common ratio of $3$,the sum of the $n$ terms is $364$ and the last term is $243$. Find the value of $n$.

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 $G.P.$

The greatest positive integer $k,$ for which $49^k+1$ is a factor of the sum $49^{125}+49^{124}+\ldots+49^{2}+49+1$ is

The least positive integer $n$ such that $1 - \frac{2}{3} - \frac{2}{3^2} - \dots - \frac{2}{3^{n-1}} < \frac{1}{100}$ 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 $[r]$ denotes the greatest integer less than or equal to $r$,then $3[r] + [-r]$ is equal to:

The numbers $(\sqrt{2} + 1), 1, (\sqrt{2} - 1)$ will be in