If the sum of $n$ terms of an $A.P.$ is $3n^2 + 5n$ and $T_m = 164$,then $m = $

The sum of $100$ terms of the series $0.9 + 0.09 + 0.009 + \dots$ will be

$11^3 + 12^3 + .... + 20^3$

The sum of $1 + 3 + 5 + 7 + \dots$ up to $n$ terms is:

For any odd integer $n \ge 1$,${n^3} - {(n - 1)^3} + ........... + {( - 1)^{n - 1}}{1^3} = $

If the sum of the first $15$ terms of the series ${\left( {\frac{3}{4}} \right)^3} + {\left( {1\frac{1}{2}} \right)^3} + {\left( {2\frac{1}{4}} \right)^3} + {3^3} + {\left( {3\frac{3}{4}} \right)^3} + \dots$ is equal to $225k$,then $k$ is equal to