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:

Let $a, b$ and $c$ be in $G.P.$ with common ratio $r,$ where $a \ne 0$ and $0 < r \le \frac{1}{2}.$ If $3a, 7b$ and $15c$ are the first three terms of an $A.P.,$ then the $4^{th}$ term of this $A.P.$ is

The value of $\left[\frac{2^{2020}+1}{2^{2018}+1}\right]+\left[\frac{3^{2020}+1}{3^{2018}+1}\right]+\left[\frac{4^{2020}+1}{4^{2018}+1}\right] +\left[\frac{5^{2020}+1}{5^{2018}+1}\right] + \left[\frac{6^{2020}+1}{6^{2018}+1}\right]$ is (where $[\cdot]$ denotes the greatest integer function):

Write the first five terms of the sequence defined by $a_{1} = -1$ and $a_{n} = \frac{a_{n-1}}{n}$ for $n \geq 2$,and obtain the corresponding series.

If $t_n$ is the $n^{th}$ term of an arithmetic progression and $t_7 = 9$,what is the value of the common difference $d$ that minimizes the product $t_1 t_2 t_7$?

If the integers from $1$ to $2021$ are written as a single integer like $123 \dots 91011 \dots 20202021$,then the $2021^{st}$ digit (counted from the left) in the resulting number is