Maximum value of sum of arithmetic progression $50, 48, 46, 44 ........$ is :-

Let $l_1, l_2, \ldots, l_{100}$ be consecutive terms of an arithmetic progression with common difference $d_1$, and let $w_1, w_2, \ldots, w_{100}$ be consecutive terms of another arithmetic progression with common difference $d_2$, where $d_1 d_2=10$. For each $i=1,2, \ldots, 100$, let $R_i$ be a rectangle with length $l_i$, width $w_i$ and area $A_i$. If $A_{51}-A_{50}=1000$, then the value of $A_{100}-A_{90}$ is. . . . . 

If the sum of $n$ terms of an $A.P.$ is $\left(p n+q n^{2}\right),$ where $p$ and $q$ are constants, find the common difference.

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

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

If $^n{C_4},{\,^n}{C_5},$ and ${\,^n}{C_6},$ are in $A.P.,$ then $n$ can be