Write the first five terms of the following sequence and obtain the corresponding series:
$a_{1} = a_{2} = 2, a_{n} = a_{n-1} - 1, n > 2$

If $\log _e a, \log _e b, \log _e c$ are in an $A.P.$ and $\log _e a - \log _e 2b, \log _e 2b - \log _e 3c, \log _e 3c - \log _e a$ are also in an $A.P.$,then $a : b : c$ is equal to

Three numbers form a $G.P.$ If the $3^{rd}$ term is decreased by $64$,the three numbers thus obtained constitute an $A.P.$ If the second term of this $A.P.$ is decreased by $8$,a $G.P.$ is formed again. Find the numbers.

If the sum of the first $40$ terms of the series $3+4+8+9+13+14+18+19+\ldots$ is $(102)m$,then $m$ is equal to:

Let $a_n = \frac{10^n}{n!}$ for $n = 1, 2, 3, \ldots$. The greatest value of $n$ for which $a_n$ is the greatest is:

If $a_1 = a_2 = 2$ and $a_n = a_{n-1} - 1$ for $n > 2$,then $a_5$ is