If the sum of the first $2n$ terms of $2, 5, 8, \dots$ is equal to the sum of the first $n$ terms of $57, 59, 61, \dots$, then $n$ is equal to

$1 + \frac{1^3 + 2^3}{1 + 2} + \frac{1^3 + 2^3 + 3^3}{1 + 2 + 3} + \dots + \frac{1^3 + 2^3 + 3^3 + \dots + 15^3}{1 + 2 + 3 + \dots + 15} - \frac{1}{2}(1 + 2 + 3 + \dots + 15)$ is equal to

If the sum of two extreme numbers of an $A.P.$ with four terms is $8$ and the product of the remaining two middle terms is $15$, then the greatest number of the series will be:

If $^nC_4, ^nC_5,$ and $^nC_6$ are in $A.P.,$ then $n$ can 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 $a_1^2 + a_2^2 + \dots + a_{17}^2 = 140m$,then $m = \dots$

The sum of the series $1^2 + 2(2^2) + 3^2 + 2(4^2) + 5^2 + 2(6^2) + \dots + 2(2m)^2$ is