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

Suppose the sequence $a_1, a_2, a_3, \ldots$ is an arithmetic progression of distinct numbers such that the sequence $a_1, a_2, a_4, a_8, \ldots$ is a geometric progression. The common ratio of the geometric progression is

If $a$ is the arithmetic mean of $b$ and $c$ and $G_1, G_2$ are the two geometric means between them,then $G_1^3 + G_2^3 = $

Let $\{a_k\}$ and $\{b_k\}, k \in N$,be two $G$.$P$.s with common ratios $r_1$ and $r_2$ respectively such that $a_1=b_1=4$ and $r_1 < r_2$. Let $c_k=a_k+b_k, k \in N$. If $c_2=5$ and $c_3=13/4$,then $\sum_{k=1}^{\infty} c_k - (12a_6 + 8b_4)$ is equal to

If for $x, y \in \mathbb{R}, x > 0,$ $y = \log_{10} x + \log_{10} x^{1/3} + \log_{10} x^{1/9} + \dots$ up to $\infty$ terms and $\frac{2+4+6+\dots+2y}{3+6+9+\dots+3y} = \frac{4}{\log_{10} x}$,then the ordered pair $(x, y)$ is equal to:

The sum $1 \cdot 1^2 - 2 \cdot 3^2 + 3 \cdot 5^2 - 4 \cdot 7^2 + 5 \cdot 9^2 - \ldots + 15 \cdot 29^2$ is $.......$.