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

If $p$ times the $p^{th}$ term of an $A.P.$ is equal to $q$ times the $q^{th}$ term of an $A.P.$,then the $(p + q)^{th}$ term is

Let $a, b$ and $c$ be the $7^{th}, 11^{th}$ and $13^{th}$ terms respectively of a non-constant $A.P.$ If these are also the three consecutive terms of a $G.P.$,then $\frac{a}{c}$ is equal to

If $a_1, a_2, a_3, .... a_{21}$ are in $A.P.$ and $a_3 + a_5 + a_{11} + a_{17} + a_{19} = 10$,then the value of $\sum_{r=1}^{21} a_r$ is:

The sum of the series $\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{n^2 - 1} + \sqrt{n^2}}$ equals

Write the seventh term of the following $GP$: $6, 12, 24, 48, \ldots$