If $\frac{a + bx}{a - bx} = \frac{b + cx}{b - cx} = \frac{c + dx}{c - dx}, (x \ne 0)$,then $a, b, c, d$ are in

If $T_n = (n^2 + 1)n!$ and $S_n = T_1 + T_2 + T_3 + ...... + T_n$. Let $\frac{T_{10}}{S_{10}} = \frac{a}{b}$,where $a$ and $b$ are relatively prime natural numbers,then the value of $(b - a)$ is:

If $a, b, c, d$ are positive real numbers such that $a + b + c + d = 2$,then $M = (a + b)(c + d)$ satisfies the relation

Let $a, b, c, d \in R^+$ such that $256 abcd \geq (a+b+c+d)^4$ and $3a + b + 2c + 5d = 11$. Then,the value of $a^3 + b + c^2 + 5d$ is equal to:

$\left[\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\cdots+\frac{1}{99 \times 100}\right]$ is equal to

If $S_n = 1 + \frac{1}{2} + \frac{1}{2^2} + \dots + \frac{1}{2^{n-1}}$,then the least integral value of $n$ such that $2 - S_n < \frac{1}{100}$ is