Let $T_r$ be the $r^{\text{th}}$ term of an $A.P.$ If for some $m$,$T_m = \frac{1}{25}$,$T_{25} = \frac{1}{20}$ and $20 \sum_{r=1}^{25} T_r = 13$,then $5m \sum_{r=m}^{2m} T_r$ is equal to:

$A$ man counts $4500$ currency notes. Let $a_n$ denote the number of notes he counts in the $n^{th}$ minute. If $a_1 = a_2 = \dots = a_{10} = 150$ and $a_{10}, a_{11}, \dots$ form an arithmetic progression with a common difference of $-2$,then how many minutes will it take for him to count all the notes?

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

Let $S_n$ denote the sum of $n$ terms of an $A.P.$ If $S_{2n} = 3S_n$,then the ratio $\frac{S_{3n}}{S_n} = $

If $\frac{a^{n}+b^{n}}{a^{n-1}+b^{n-1}}$ is the $A.M.$ between $a$ and $b,$ then find the value of $n$.

The number of terms of the $A.P. 3, 7, 11, 15, ...$ to be taken so that the sum is $406$ is