Consider an A.P.: $a_1, a_2, \ldots, a_{\mathrm{n}} ; a_1>0$. If $a_2-a_1=\frac{-3}{4}, a_{\mathrm{n}}=\frac{1}{4} a_1$, and $\sum\limits_{\mathrm{i}=1}^{\mathrm{n}} a_{\mathrm{i}}=\frac{525}{2}$, then $\sum\limits_{\mathrm{i}=1}^{17} a_{\mathrm{i}}$ is equal to

Let $\sum\limits_{k=1}^n a_k=\alpha n^2+\beta n$. If $a_{10}=59$ and $a_6=7 a_1$, then $\alpha+\beta$ is equal to :

If the sum of the first four terms of an A.P. is 6 and the sum of its first six terms is 4 , then the sum of its first twelve terms is

The positive integer n, for which the solutions of the equation

$x(x+2) + (x+2)(x+4) + \cdots + (x+2n-2)(x+2n) = \frac{8n}{3}$ are two consecutive even integers, is :