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

Let $C_0$ be a circle of radius $1$. For $n \geq 1$,let $C_n$ be a circle whose area equals the area of a square inscribed in $C_{n-1}$. Then,$\sum_{i=0}^{\infty} \text{Area}(C_i)$ equals

Let $A_n = \left( \frac{3}{4} \right) - \left( \frac{3}{4} \right)^2 + \left( \frac{3}{4} \right)^3 - \dots + (-1)^{n-1} \left( \frac{3}{4} \right)^n$ and $B_n = 1 - A_n$. Then,the least odd natural number $p$ such that $B_n > A_n$ for all $n \geq p$ is:

If the product of three terms of a $G.P.$ is $512$. If $8$ is added to the first term and $6$ is added to the second term,the resulting numbers are in $A.P.$. Find the numbers.

In a geometric progression,if the ratio of the sum of the first $5$ terms to the sum of their reciprocals is $49$,and the sum of the first and the third term is $35$,then the first term of this geometric progression is:

Show that the products of the corresponding terms of the sequences $a, ar, ar^{2}, \dots, ar^{n-1}$ and $A, AR, AR^{2}, \dots, AR^{n-1}$ form a $G.P.$,and find the common ratio.