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

Let $a$ and $b$ be two distinct positive real numbers. Let the $11^{\text{th}}$ term of a $GP$,whose first term is $a$ and third term is $b$,be equal to the $p^{\text{th}}$ term of another $GP$,whose first term is $a$ and fifth term is $b$. Then $p$ is equal to:

If $\frac{a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - dx}$ and $x \neq 0$,then $a, b, c, d$ are in

$A$ person writes a letter to four of his friends. He asks each one of them to copy the letter and mail it to four different persons with the instruction that they continue the chain similarly. Assuming that the chain is not broken and that it costs $50$ paise to mail one letter,find the total amount spent on postage when the $8^{\text{th}}$ set of letters is mailed.

Find the sum of the first $n$ terms and the sum of the first $5$ terms of the geometric series $1 + \frac{2}{3} + \frac{4}{9} + \dots$

$\frac{6}{3^{26}} + \frac{10 \cdot 1}{3^{25}} + \frac{10 \cdot 2}{3^{24}} + \frac{10 \cdot 2^2}{3^{23}} + \ldots + \frac{10 \cdot 2^{24}}{3}$ is equal to