If $x_1, x_2, \dots, x_n$ and $\frac{1}{h_1}, \frac{1}{h_2}, \dots, \frac{1}{h_n}$ are two $A.P.s$ such that $x_3 = h_2 = 8$ and $x_8 = h_7 = 20$,then $x_5 \cdot h_{10}$ equals

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

The $G.M.$ of the numbers $3, 3^2, 3^3, ..., 3^n$ is

The number of terms of the $A.P. 3, 7, 11, 15, ...$ to be taken so that the sum is $406$ is

If twice the $11^{th}$ term of an $A.P.$ is equal to $7$ times of its $21^{st}$ term,then its $25^{th}$ term is equal to

The sum of the following series $1 + 6 + \frac{9(1^2 + 2^2 + 3^2)}{7} + \frac{12(1^2 + 2^2 + 3^2 + 4^2)}{9} + \frac{15(1^2 + 2^2 + ... + 5^2)}{11} + ...$ up to $15$ terms is: