Let ${S_n} = \frac{1}{{{1^3}}} + \frac{{1 + 2}}{{{1^3} + {2^3}}} + \frac{{1 + 2 + 3}}{{{1^3} + {2^3} + {3^3}}} + \dots + \frac{{1 + 2 + \dots + n}}{{{1^3} + {2^3} + \dots + {n^3}}}$. If $100 S_n = n$,then $n$ is equal to:

Let $A_n = \left( \frac{3}{4} \right) - \left( \frac{3}{4} \right)^2 + \left( \frac{3}{4} \right)^3 - \dots + (-1)^{n-1} \left( \frac{3}{4} \right)^n$ and $B_n = 1 - A_n$. Then,the least odd natural number $p$,such that $B_n > A_n$ for all $n \geq p$,is

Consider two positive numbers $a$ and $b$. If the arithmetic mean of $a$ and $b$ exceeds their geometric mean by $\frac{3}{2}$ and the geometric mean of $a$ and $b$ exceeds their harmonic mean by $\frac{6}{5}$,then the absolute value of $(a^2 - b^2)$ is equal to

The arithmetic mean of the first $n$ natural numbers is:

The number $111...1$ ($91$ times) is a:

If the roots of the equation ${x^3} - 12{x^2} + 39x - 28 = 0$ are in $A.P.$,then their common difference will be