What is the sum of all two-digit numbers which give a remainder of $4$ when divided by $6$?

Let $a, b, c \in R$. If $f(x) = ax^2 + bx + c$ is such that $a + b + c = 3$ and $f(x + y) = f(x) + f(y) + xy$ for all $x, y \in R$,then $\sum_{n=1}^{10} f(n)$ is equal to:

If $a_m$ denotes the $m^{th}$ term of an $A.P.$,then $a_m$ =

Consider two positive numbers $a$ and $b$. If the arithmetic mean of $a$ and $b$ exceeds their geometric mean by $\frac{3}{2}$ and the geometric mean of $a$ and $b$ exceeds their harmonic mean by $\frac{6}{5}$,then the absolute value of $(a^2 - b^2)$ is equal to

If $m$ is the $A.M.$ of two distinct real numbers $l$ and $n$ $(l, n > 1)$ and $G_1, G_2,$ and $G_3$ are three geometric means between $l$ and $n$,then $G_1^4 + 2G_2^4 + G_3^4$ equals:

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