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
$2^{\frac{1}{2}}$
$2^{\frac{1}{4}}$
$2$
$1$
If the sum of $n$ terms of a $G.P.$ is $255$ and ${n^{th}}$ terms is $128$ and common ratio is $2$, then first term will be
If $a _{1}(>0), a _{2}, a _{3}, a _{4}, a _{5}$ are in a G.P., $a _{2}+ a _{4}=2 a _{3}+1$ and $3 a _{2}+ a _{3}=2 a _{4}$, then $a _{2}+ a _{4}+2 a _{5}$ is equal to
The $G.M.$ of roots of the equation ${x^2} - 18x + 9 = 0$ is
If three successive terms of a$G.P.$ with common ratio $r(r>1)$ are the lengths of the sides of a triangle and $[\mathrm{r}]$ denotes the greatest integer less than or equal to $r$, then $3[r]+[-r]$ is equal to :
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
$a^{q-r} b^{r-p} c^{p-q}=1$