$\sum\limits_{m = 1}^n {{m^2}}$ is equal to

If the $7^{th}$ term of a $H.P.$ is $\frac{1}{10}$ and the $12^{th}$ term is $\frac{1}{25}$,then the $20^{th}$ term is

The ratio of the sum of $m$ and $n$ terms of an $A.P.$ is $m^2 : n^2$. Then the ratio of the $m^{th}$ and $n^{th}$ term will be:

The value of $(2 \cdot {}^{1}P_{0} - 3 \cdot {}^{2}P_{1} + 4 \cdot {}^{3}P_{2} - \dots$ up to $51^{\text{th}}$ term) + $(1! - 2! + 3! - \dots$ up to $51^{\text{th}}$ term) is equal to

If the sides of a right-angled triangle are in $A.P.$,then the sides are proportional to

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