Find the sum of all two-digit numbers which,when divided by $4$,yield $1$ as a remainder.

The ratio of the sums of $m$ and $n$ terms of an $A.P.$ is $m^{2}: n^{2}$. Show that the ratio of the $m^{th}$ and $n^{th}$ term is $(2m-1):(2n-1)$.

If the $p^{th}$,$q^{th}$,and $r^{th}$ terms of an Arithmetic Progression are $a$,$b$,and $c$ respectively,then $[a(q - r) + b(r - p) + c(p - q)] = ?$

If the sum of first $n$ terms of an $A.P.$ is equal to the sum of its first $m$ terms,$(m \ne n)$,then the sum of its first $(m + n)$ terms will be

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 $m = \dots$

Let $S_{1}$ be the sum of the first $2n$ terms of an arithmetic progression. Let $S_{2}$ be the sum of the first $4n$ terms of the same arithmetic progression. If $(S_{2} - S_{1})$ is $1000$,then the sum of the first $6n$ terms of the arithmetic progression is equal to: