If the sum of an infinite geometric series is $\frac{4}{5}$ and its $1^{st}$ term is $\frac{3}{4}$, then its common ratio is

If there are $5$ geometric means between $486$ and $\frac{2}{3}$,what is the $4^{th}$ geometric mean?

Let $a, b, c, d$ and $p$ be any non-zero distinct real numbers such that $(a^{2}+b^{2}+c^{2}) p^{2} - 2(ab+bc+cd) p + (b^{2}+c^{2}+d^{2}) = 0$. Then:

Find the geometric mean of the first $n$ natural numbers.

If $r > 1$,$x = a + \frac{a}{r} + \frac{a}{r^2} + \dots \infty$,$y = b - \frac{b}{r} + \frac{b}{r^2} - \dots \infty$,and $z = c + \frac{c}{r^2} + \frac{c}{r^4} + \dots \infty$,then $\frac{xy}{z} = \dots$

The sum of the first two terms of a geometric progression is $12$. The sum of the third and fourth terms is $48$. The terms of the geometric progression alternate between positive and negative. What is the first term?