If $1, \log_9(3^{1-x} + 2), \log_3(4 \cdot 3^x - 1)$ are in an arithmetic progression,then $x = \dots$

Suppose the sum of the first $m$ terms of an arithmetic progression is $n$ and the sum of its first $n$ terms is $m$,where $m \neq n$. Then,the sum of the first $(m+n)$ terms of the arithmetic progression is

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:

The minimum value of ${\left( {\frac{3}{a} - 1} \right)^2} + {\left( {\frac{a}{b} - 1} \right)^2} + {\left( {\frac{b}{c} - 1} \right)^2} + {\left( {3c - 1} \right)^2}$ where $0 < a, b, c \leqslant 9$,is $p - q\sqrt{r}$; $p, q, r \in I$ and $q, r$ are coprime,then $(p + q + r)$ is equal to

Which term of the sequence $5, 8, 11, 14, \dots$ is $320$?

Let $a_{1}, a_{2}, a_{3}, \ldots$ be an $A.P.$ If $\frac{a_{1}+a_{2}+\ldots+a_{10}}{a_{1}+a_{2}+\ldots+a_{p}}=\frac{100}{p^{2}}, p \neq 10$,then $\frac{a_{11}}{a_{10}}$ is equal to :