If the variance of the terms in an increasing $A.P.$,$b_{1}, b_{2}, b_{3}, \ldots, b_{11}$ is $90$,then the common difference of this $A.P.$ is

$A$ person has two parents (father and mother),four grandparents,eight great-grandparents,and so on. Find the total number of ancestors the person has up to the $10^{th}$ generation.

If $y = x + x^2 + x^3 + \dots \infty$,then $x = $

The sum to infinity of the following series $2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{2^3} + \frac{1}{3^3} + \dots$ will be

If $\frac{1}{p + q}, \frac{1}{r + p}, \frac{1}{q + r}$ are in $A.P.$,then

For a series $S = 1 - 2 + 3 - 4 + \dots$ up to $n$ terms,
Statement $-1$: The sum of the series is always dependent on the value of $n$,i.e.,whether it is even or odd.
Statement $-2$: The sum of the series is $-\frac{n}{2}$ when $n$ is any even integer.