For every natural number $n$,$n(n + 1)$ is always

If $\log _e a, \log _e b, \log _e c$ are in an $A.P.$ and $\log _e a - \log _e 2b, \log _e 2b - \log _e 3c, \log _e 3c - \log _e a$ are also in an $A.P.$,then $a : b : c$ 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 $.......$.

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:

If $a, b, c$ are in $A.P.$,then $\frac{a}{bc}, \frac{1}{c}, \frac{2}{b}$ are in

For a positive integer $n$,let $a(n) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{2^n - 1}$. Then: