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

Let $A_{1}, A_{2}, A_{3}, \ldots$ be squares such that for each $n \geq 1,$ the length of the side of $A_{n}$ equals the length of the diagonal of $A_{n+1}$. If the side length of $A_{1}$ is $12 \text{ cm}$,then the smallest value of $n$ for which the area of $A_{n}$ is less than $1 \text{ cm}^2$ is:

The roots of the equation $x^5 - 40x^4 + px^3 + qx^2 + rx + s = 0$ are in $G.P.$ The sum of their reciprocals is $10$. Then the value of $|s|$ is

The number $111...1$ ($91$ times) is a

Divide $155$ into three parts such that the three numbers are in a Geometric Progression $(GP)$ and the first term is $120$ less than the third term.

If we insert two numbers between $\sqrt{2}$ and $4$ so that the resulting sequence is in $G.P.$, then the inserted numbers in the order are