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

Let $a_{n}$ be the $n^{\text{th}}$ term of a $G$.$P$. of positive terms. If $\sum_{n=1}^{100} a_{2n+1} = 200$ and $\sum_{n=1}^{100} a_{2n} = 100$,then $\sum_{n=1}^{200} a_{n}$ is equal to:

If the arithmetic mean of two numbers $a$ and $b$,$a > b > 0$,is five times their geometric mean,then $\frac{a + b}{a - b}$ is equal to

If $t_n = \frac{1}{4}(n + 2)(n + 3)$ for $n = 1, 2, 3, \dots$,then $\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} + \dots + \frac{1}{t_{2003}} = $

If $ < a_n > $ is an $A.P.$ and $a_1 + a_4 + a_7 + .......+ a_{16} = 147$, then $a_1 + a_6 + a_{11} + a_{16}$ is equal to

How many terms are there in an $A.P.$ whose first and fifth terms are $-14$ and $2,$ respectively,and the sum of terms is $40$?