Let ${a_1}, {a_2}, \dots, {a_{49}}$ be in $A.P.$ such that $\sum_{k = 0}^{12} {a_{4k + 1}} = 416$ and ${a_9} + {a_{43}} = 66$. If $\sum_{r = 1}^{17} a_r^2 = 140m$,then $m = \dots$

If $x, y, z \in \mathbb{R}^+$ such that $x + y + z = 4$,then the maximum possible value of $xyz^2$ is

If ${a_1}, {a_2}, {a_3}, \dots, {a_{24}}$ are in arithmetic progression and ${a_1} + {a_5} + {a_{10}} + {a_{15}} + {a_{20}} + {a_{24}} = 225$,then ${a_1} + {a_2} + {a_3} + \dots + {a_{23}} + {a_{24}} = $

Consider an $A.P.$ of positive integers,whose sum of the first three terms is $54$ and the sum of the first twenty terms lies between $1600$ and $1800$. Then its $11^{\text{th}}$ term is:

In an $A.P.$,if the $m^{\text{th}}$ term is $n$ and the $n^{\text{th}}$ term is $m$,where $m \neq n$,find the $p^{\text{th}}$ term.

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