The product of three geometric means between $4$ and $\frac{1}{4}$ will be

The $5^{\text{th}}$,$8^{\text{th}}$,and $11^{\text{th}}$ terms of a $G.P.$ are $p, q$,and $s$,respectively. Show that $q^{2} = ps$.

If the roots of $x^3-k x^2+14 x-8=0$ are in geometric progression,then $k$ is equal to

The coefficient of $x^{49}$ in the expansion of $(x - 1)(x - \frac{1}{2})(x - \frac{1}{2^2}) \dots (x - \frac{1}{2^{49}})$ is equal to

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:

For a $G.P.$,if $(m+n)^{\text{th}}$ term is $p$ and $(m-n)^{\text{th}}$ term is $q$,then the $m^{\text{th}}$ term is $.........$