An arithmetic progression is written in the following way. The sum of all the terms of the $10^{\text{th}}$ row is..........

Let four distinct positive numbers $a_1, a_2, a_3, a_4$ be in a geometric progression. Let $b_1 = a_1$,$b_2 = b_1 + a_2$,$b_3 = b_2 + a_3$,and $b_4 = b_3 + a_4$.
Statement-$I$: The numbers $b_1, b_2, b_3, b_4$ are neither in an arithmetic progression nor in a geometric progression.
Statement-$II$: The numbers $b_1, b_2, b_3, b_4$ are in a harmonic progression.

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

The sum $1 \cdot 1^2 - 2 \cdot 3^2 + 3 \cdot 5^2 - 4 \cdot 7^2 + 5 \cdot 9^2 - \ldots + 15 \cdot 29^2$ is $.......$.

Let $a, b, c > 1$. If $a^3, b^3, c^3$ are in $A.P.$ and $\log_a b, \log_c a, \log_b c$ are in $G.P.$,and the sum of the first $20$ terms of an $A.P.$ with first term $\frac{a+4b+c}{3}$ and common difference $\frac{a-8b+c}{10}$ is $-444$,then $abc$ 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