Which term of the sequence $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \dots$ is $\frac{1}{19683}$?

All terms of a geometric progression are positive. If each term is equal to the sum of the two terms following it,then the common ratio of this progression is:

If $a, b, c$ and $d$ are in $G.P.$,show that:
$(a^{2}+b^{2}+c^{2})(b^{2}+c^{2}+d^{2})=(ab+bc+cd)^{2}$

The arithmetic mean of two numbers $b$ and $c$ is $a$,and $g_1$ and $g_2$ are two geometric means between them. If $g_1^3 + g_2^3 = kabc$,then $k = \dots$

The number which should be added to the numbers $2, 14, 62$ so that the resulting numbers may be in $G.P.$ is

If the sum of an infinite $GP$ $a, ar, ar^{2}, ar^{3}, \ldots$ is $15$ and the sum of the squares of its each term is $150$,then the sum of $ar^{2}, ar^{4}, ar^{6}, \ldots$ is: