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

Let $a, b \in \mathbb{R}$ be such that $a, a + 2b, 2a + b$ are in $A.P.$ and $(b + 1)^2, ab + 5, (a + 1)^2$ are in $G.P.$ Then $(a + b)$ equals:

If $\frac{b + a}{b - a} = \frac{b + c}{b - c}$,then $a, b, c$ are in

The sum of three consecutive terms in a geometric progression is $14$. If $1$ is added to the first and the second terms and $1$ is subtracted from the third,the resulting new terms are in arithmetic progression. Then the lowest of the original terms is

Let $a_i = i + \frac{1}{i}$ for $i = 1, 2, \ldots, 20$. Let $p = \frac{1}{20} \sum_{i=1}^{20} a_i$ and $q = \frac{1}{20} \sum_{i=1}^{20} \frac{1}{a_i}$. Then,

If $|x| < 1, |y| < 1$ and $x \neq y,$ then the sum to infinity of the following series $(x+y)+(x^{2}+xy+y^{2})+(x^{3}+x^{2}y+xy^{2}+y^{3})+\ldots$ is: