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

If $a, b, c$ are in $A.P.$,then $\frac{1}{bc}, \frac{1}{ca}, \frac{1}{ab}$ will be in

Suppose that all the terms of an arithmetic progression $(A.P.)$ are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is $6:11$ and the seventh term lies between $130$ and $140$,then the common difference of this $A.P.$ is

If $\log_3 2, \log_3(2^x - 5)$ and $\log_3(2^x - \frac{7}{2})$ are in $A.P.$,then $x$ is equal to

If the sum of $n$ terms of an arithmetic progression is $2n^2 + 5n$,then its $n^{th}$ term is.........

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 $m = \dots$