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 :

Let $a_1, \frac{a_2}{2}, \frac{a_3}{2^2}, \ldots, \frac{a_{10}}{2^9}$ be a G.P. of common ratio $\frac{1}{\sqrt{2}}$. If $a_1 + a_2 + \ldots + a_{10} = 62$, then $a_1$ is equal to:

Let $a_1, a_2, a_3, \ldots$ be a G.P. of increasing positive terms such that $a_2 \cdot a_3 \cdot a_4=64$ and $a_1+a_3+a_5=\frac{813}{7}$. Then $a_3+a_5+a_7$ is equal to :