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:

Let the sequence $a_1, a_2, a_3, \dots, a_{2n}$ form an $A.P.$ Then $a_1^2 - a_2^2 + a_3^2 - a_4^2 + \dots + a_{2n - 1}^2 - a_{2n}^2 = $

If the first term of an arithmetic progression is $2$ and the common difference is $4$,then the sum of its first $40$ terms is........

The maximum sum of the series $20 + 19\frac{1}{3} + 18\frac{2}{3} + \dots$ is

For $p, q \in R$,consider the real-valued function $f(x) = (x - p)^2 - q$,where $x \in R$ and $q > 0$. Let $a_1, a_2, a_3, a_4$ be in an arithmetic progression with mean $p$ and a positive common difference $d$. If $|f(a_i)| = 500$ for all $i = 1, 2, 3, 4$,then the absolute difference between the roots of $f(x) = 0$ is:

Let $a_1, a_2, a_3, \ldots, a_{11}$ be real numbers satisfying $a_1=15$,$27-2a_2 > 0$,and $a_k = 2a_{k-1} - a_{k-2}$ for $k = 3, 4, \ldots, 11$. If $\frac{a_1^2 + a_2^2 + \ldots + a_{11}^2}{11} = 90$,then the value of $\frac{a_1 + a_2 + \ldots + a_{11}}{11}$ is equal to