The third term of a $G.P.$ is the square of the first term. If the second term is $8$,then the $6^{th}$ term is:

If the first and the $n^{\text{th}}$ term of a $G.P.$ are $a$ and $b$ respectively,and if $P$ is the product of $n$ terms,prove that $P^{2} = (ab)^{n}$.

Find the $20^{\text{th}}$ and $n^{\text{th}}$ terms of the $G.P.$ $\frac{5}{2}, \frac{5}{4}, \frac{5}{8}, \dots$

Let $a, ar, ar^2, \ldots$ be an infinite $G.P.$ If $\sum_{n=0}^{\infty} ar^n = 57$ and $\sum_{n=0}^{\infty} a^3 r^{3n} = 9747$,then $a + 18r$ is equal to:

If $\frac{1}{2 \cdot 3^{10}}+\frac{1}{2^{2} \cdot 3^{9}}+\ldots+\frac{1}{2^{10} \cdot 3}=\frac{K}{2^{10} \cdot 3^{10}}$,then the remainder when $K$ is divided by $6$ is

$\frac{6}{3^{26}} + \frac{10 \cdot 1}{3^{25}} + \frac{10 \cdot 2}{3^{24}} + \frac{10 \cdot 2^2}{3^{23}} + \ldots + \frac{10 \cdot 2^{24}}{3}$ is equal to