The sum of the first $20$ terms of the sequence $0.7, 0.77, 0.777, \dots$ is

Which term of the geometric sequence $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots$ is $\frac{1}{19683}$?

If $x_1, x_2, \dots, x_n$ and $\frac{1}{h_1}, \frac{1}{h_2}, \dots, \frac{1}{h_n}$ are two $A.P.s$ such that $x_3 = h_2 = 8$ and $x_8 = h_7 = 20$,then $x_5 \cdot h_{10}$ equals

The sum of $n$ terms of the following series $1 + (1 + x) + (1 + x + x^2) + \dots$ will be

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

Four numbers are in arithmetic progression. The sum of the first and last term is $8$ and the product of both middle terms is $15$. The least number of the series is