Find the sum of infinite terms of the series $1+\frac{2}{3}+\frac{3}{3^{2}}+\frac{4}{3^{3}}+\frac{5}{3^{4}}+\cdots$

If the sum of the first $6$ terms is $9$ times the sum of the first $3$ terms of the same $G.P.$,then the common ratio of the series will be

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

The sum $1(1!) + 2(2!) + 3(3!) + ....+n (n!)$ equals

Let $a_{1}, a_{2}, a_{3}, \ldots$ be a $G$.$P$. such that $a_{1} < 0$; $a_{1} + a_{2} = 4$ and $a_{3} + a_{4} = 16$. If $\sum_{i=1}^{9} a_{i} = 4 \lambda$,then $\lambda$ is equal to:

The sum of the first three terms of a $G.P.$ is $S$ and their product is $27$. Then all such $S$ lie in