What is the sum of the first $40$ terms of the series $1 + 2 + 3 + 4 + 5 + 8 + 7 + 16 + 9 + \dots$?

The product of three consecutive terms of a $G.P.$ is $512$. If $4$ is added to each of the first and the second of these terms,the three terms now form an $A.P.$ Then the sum of the original three terms of the given $G.P.$ is

Let $S_{n}(x) = \log_{a^{1/2}} x + \log_{a^{1/3}} x + \log_{a^{1/6}} x + \log_{a^{1/11}} x + \log_{a^{1/18}} x + \log_{a^{1/27}} x + \ldots$ up to $n$-terms,where $a > 1$. If $S_{24}(x) = 1093$ and $S_{12}(2x) = 265$,then the value of $a$ is equal to ..... .

Let $a_{1}=1$ and for $n \ge 1$, $a_{n+1} = \frac{1}{2}a_{n} + \frac{n^{2}-2n-1}{n^{2}(n+1)^{2}}$. Then $|\sum_{n=1}^{\infty}(a_{n}-\frac{2}{n^{2}})|$ is equal to ........... .

Let $a_0=0$ and $a_n=3 a_{n-1}+1$ for $n \geq 1$. Then,the remainder obtained by dividing $a_{2010}$ by $11$ is

Let $A_1$ and $A_2$ be two arithmetic means and $G_1, G_2, G_3$ be three geometric means between two distinct positive numbers $a$ and $b$. Then $G_1^4 + G_2^4 + G_3^4 + G_1^2 G_3^2$ is equal to