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

Consider the quadratic equation $(n^2 - 2n + 2)x^2 - 3x + (n^2 - 2n + 2) = 0, n \in R$. Let $\alpha$ be the minimum value of the product of its roots and $\beta$ be the maximum value of the sum of its roots. Then the sum of the first six terms of the $G$.$P$.,whose first term is $\alpha$ and the common ratio is $\frac{\alpha}{\beta}$, is:

Let $a_1, \frac{a_2}{2}, \frac{a_3}{2^2}, \ldots, \frac{a_{10}}{2^9}$ be a $G$.$P$. with common ratio $r = \frac{1}{\sqrt{2}}$. If $a_1 + a_2 + \ldots + a_{10} = 62$, then $a_1$ is equal to:

Let $x_1, x_2, \ldots, x_6$ be the roots of the polynomial equation $x^6+2 x^5+4 x^4+8 x^3+16 x^2+32 x+64=0$. Then,

If $x > 1, y > 1, z > 1$ are in geometric progression,then in which progression are $\frac{1}{1 + \ln x}, \frac{1}{1 + \ln y}, \frac{1}{1 + \ln z}$?

If $\frac{a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - dx}$ and $x \neq 0$,then $a, b, c$ and $d$ are in: