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

The arithmetic mean of the first $n$ natural numbers is:

The $4$th term of an $A.P.$ is equal to $3$ times the first term and the seventh term exceeds twice the third term by $1$. Find the first term and the common difference.

The sum of all terms of the $n^{th}$ bracket of the sequence $(1), (3, 5), (7, 9, 11), \dots$ is equal to:

Suppose that a function $f: R \rightarrow R$ satisfies $f(x+y)=f(x) f(y)$ for all $x, y \in R$ and $f(1)=3$. If $\sum_{i=1}^{n} f(i)=363$,then $n$ is equal to

The number of terms in an $A.P.$ is even. The sum of the odd terms is $24$ and the sum of the even terms is $30$. If the last term exceeds the first term by $10\frac{1}{2}$,then the number of terms in the $A.P.$ is: