If the sum of $n$ terms of an $AP$ is given by $S_{n} = n^{2} + n$,then the common difference of the $AP$ 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}} = $

Given the sum of the first $n$ terms of an $A.P.$ is $S_n = 2n + 3n^2$. Another $A.P.$ is formed with the same first term and double the common difference. The sum of $n$ terms of the new $A.P.$ is:

If $\frac{a^{n}+b^{n}}{a^{n-1}+b^{n-1}}$ is the $A.M.$ between $a$ and $b,$ then find the value of $n$.

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 $m = \dots$

The common difference of the $A.P.$ $b_{1}, b_{2}, \ldots, b_{m}$ is $2$ more than the common difference of $A.P.$ $a_{1}, a_{2}, \ldots, a_{n}$. If $a_{40} = -159$,$a_{100} = -399$ and $b_{100} = a_{70}$,then $b_{1}$ is equal to: