The sum of all natural numbers between $1$ and $100$ which are multiples of $3$ is

If $\log 2, \log (2^n - 1)$ and $\log (2^n + 3)$ are in $A.P.$,then $n =$

If $a, b, c, d, e, f$ are in an arithmetic progression,then $e - c = \dots$

If $x, y, z \in R^+$ such that $x + y + z = 4$,then the maximum possible value of $xyz^2$ is -

If the sum of the first $2n$ terms of the arithmetic progression $2, 5, 8, \dots$ is equal to the sum of the first $n$ terms of the arithmetic progression $57, 59, 61, \dots$,then $n = \dots$

Let $a_1, a_2, a_3, \ldots$ be terms of an $A.P.$ If $\frac{a_1 + a_2 + \ldots + a_p}{a_1 + a_2 + \ldots + a_q} = \frac{p^2}{q^2}$ for $p \ne q$,then $\frac{a_6}{a_{21}}$ equals: