Let $a_1, a_2, \dots, a_{10}$ be in $A.P.$ and $h_1, h_2, \dots, h_{10}$ be in $H.P.$ If $a_1 = h_1 = 2$ and $a_{10} = h_{10} = 3$,then $a_4 h_7$ is

The sum of the series $1 \cdot 3^2 + 2 \cdot 5^2 + 3 \cdot 7^2 + \dots$ up to $20$ terms is

If the $p^{th}$,$q^{th}$,and $r^{th}$ terms of a $G.P.$ are $a$,$b$,and $c$ respectively,then $a^{q - r} b^{r - p} c^{p - q}$ is equal to

$\prod\limits_{n = 1}^{10} {\left( {\frac{{\left( {6\sum\limits_{i = 0}^n i } \right) + 1}}{{\left( {6\sum\limits_{j = 0}^n {(j - 1)} } \right) + 1}}} \right)} $ is equal to:

The $n^{th}$ term of the series $\frac{1^3}{1} + \frac{1^3 + 2^3}{1 + 3} + \frac{1^3 + 2^3 + 3^3}{1 + 3 + 5} + \dots$ will be

The sum $1(1!) + 2(2!) + 3(3!) + ....+n (n!)$ equals