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

Let $S_n$ denote the sum of the first $n$ terms of an $A.P$. If $S_4 = 16$ and $S_6 = -48$,then $S_{10}$ is equal to

If $\frac{S_n}{S_m} = \frac{n^4}{m^4}$ (where $S_k$ is the sum of the first $k$ terms of an $A$.$P$. $a_1, a_2, \dots$),then the value of $\frac{a_{m+1}}{a_{n+1}}$ in terms of $m$ and $n$ is:

Let $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}$ be an $A$.$P$. of four terms such that each term of the $A$.$P$. and its common difference $l$ are integers. If $\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}=48$ and $\alpha_{1}\alpha_{2}\alpha_{3}\alpha_{4}+l^{4}=361$, then the largest term of the $A$.$P$. is equal to

The sum of all natural numbers between $1$ and $100$ which are multiples of $3$ is

If the roots of the equation $4x^3 - 12x^2 + 11x + m = 0$ are in arithmetic progression,then $m =$