If $G$ is the geometric mean between $x$ and $y$,then the value of $\frac{1}{G^2 - x^2} + \frac{1}{G^2 - y^2}$ is:

If $a, b, c, d$ are in a geometric progression,then $(a^3 + b^3)^{-1}, (b^3 + c^3)^{-1}, (c^3 + d^3)^{-1}$ are in which progression?

The value of $x$ that satisfies the relation $x = 1 - x + x^2 - x^3 + x^4 - x^5 + \dots \infty$ is:

If the $p^{th}$,$q^{th}$,and $r^{th}$ terms of a geometric progression are $a, b, c$ respectively,then $a^{q-r} \cdot b^{r-p} \cdot c^{p-q} = \dots\dots$

$A$ particle starts at the origin and moves $1$ unit horizontally to the right and reaches $P_{1}$, then it moves $\frac{1}{2}$ unit vertically up and reaches $P_{2}$, then it moves $\frac{1}{4}$ unit horizontally to the right and reaches $P_{3}$, then it moves $\frac{1}{8}$ unit vertically down and reaches $P_{4}$, then it moves $\frac{1}{16}$ unit horizontally to the right and reaches $P_{5}$ and so on. Let $P_{n} = (x_{n}, y_{n})$ and $\lim_{n \rightarrow \infty} x_{n} = \alpha$ and $\lim_{n \rightarrow \infty} y_{n} = \beta$. Then, $(\alpha, \beta)$ is

If the roots of the equation $x^5-40x^4-Px^3-Rx-S=0$ are in geometric progression and the sum of the reciprocals of the roots is $10$,then $|S|=$