The sum to $n$ terms of the series $2^2 + 4^2 + 6^2 + \dots$ is

The number of terms in an $A.P.$ is even. The sum of the odd terms is $24$ and the sum of the even terms is $30$. If the last term exceeds the first term by $10\frac{1}{2}$,then the number of terms in the $A.P.$ is:

The sum of the squares of the first $n$ natural numbers exceeds their sum by $330$. Then,$n = $

$\sum\limits_{r = 0}^{100} {({r^2} + 4r + 4)(r + 1)!}$ is equal to :-

If ${A_1}, {A_2}$; ${G_1}, {G_2}$ and ${H_1}, {H_2}$ are $AMs$,$GMs$,and $HMs$ between two quantities,then the value of $\frac{{G_1 G_2}}{{H_1 H_2}}$ is

If the sum of the first $n$ terms of two arithmetic progressions $3, 7, 11, 15, \ldots$ and $30, 33, 36, 39, \ldots$ are equal,then find the value of $n$.