If $25, x - 6$,and $x - 12$ are consecutive terms of a geometric progression,then $x = \dots$

For the functions $f(\theta) = \alpha \tan^2 \theta + \beta \cot^2 \theta$ and $g(\theta) = \alpha \sin^2 \theta + \beta \cos^2 \theta$,where $\alpha > \beta > 0$,let $\min_{0 < \theta < \pi/2} f(\theta) = \max_{0 < \theta < \pi} g(\theta)$. If the first term of a $G$.$P$. is $(\frac{\alpha}{2\beta})$,its common ratio is $(\frac{2\beta}{\alpha})$ and the sum of its first $10$ terms is $\frac{m}{n}$,where $\gcd(m,n)=1$,then $m+n$ is equal to . . . . . . .

If $64, 27, 36$ are the $P^{\text{th}}$,$Q^{\text{th}}$,and $R^{\text{th}}$ terms of a $GP$,then $P+2Q$ is equal to

The arithmetic mean of the sequence $1, 2, 4, 8, 16, \ldots, 2^n$ is:

If $\frac{1}{2 \cdot 3^{10}}+\frac{1}{2^{2} \cdot 3^{9}}+\ldots+\frac{1}{2^{10} \cdot 3}=\frac{K}{2^{10} \cdot 3^{10}}$,then the remainder when $K$ is divided by $6$ is

Let $a, ar, ar^2, \ldots$ be an infinite $G.P.$ If $\sum_{n=0}^{\infty} ar^n = 57$ and $\sum_{n=0}^{\infty} a^3 r^{3n} = 9747$,then $a + 18r$ is equal to: