The product $2^{\frac{1}{4}} \cdot 4^{\frac{1}{16}} \cdot 8^{\frac{1}{48}} \cdot 16^{\frac{1}{128}} \cdot \ldots .$ to $\infty$ is equal to

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

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

If the $p^{\text {th }}, q^{\text {th }}$ and $r^{\text {th }}$ terms of a $G.P.$ are $a, b$ and $c,$ respectively. Prove that

$a^{q-r} b^{r-p} c^{p-q}=1$

The sum of two numbers is $6$ times their geometric mean, show that numbers are in the ratio $(3+2 \sqrt{2}):(3-2 \sqrt{2})$