Let $a_1, a_2, a_3, \ldots$ be a $G.P.$ of increasing positive numbers. If $a_3 a_5 = 729$ and $a_2 + a_4 = \frac{111}{4}$,then $24(a_1 + a_2 + a_3)$ is equal to

Find the sum of the first $9$ terms of the geometric series $1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \dots$

The least positive integer $n$ such that $1 - \frac{2}{3} - \frac{2}{3^2} - \dots - \frac{2}{3^{n-1}} < \frac{1}{100}$ is

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:

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.

Let $a_{1}, a_{2}, a_{3}, \dots$ be a $G$.$P$. of increasing positive terms such that $a_{2} \cdot a_{3} \cdot a_{4} = 64$ and $a_{1} + a_{3} + a_{5} = \frac{813}{7}$. Then $a_{3} + a_{5} + a_{7}$ is equal to: