If the $m^{th}$ term of an arithmetic progression is $1/n$ and the $n^{th}$ term is $1/m$,then the sum of the first $mn$ terms is:

In an $A.P.$,the sixth term is $a_6 = 2$. If the product $a_1 a_4 a_5$ is the greatest,then the common difference of the $A.P.$ is equal to

What is the sum of all integers between $81$ and $719$ that are divisible by $5$?

Let $T_r$ be the $r^{th}$ term of an $A.P.$ for $r = 1, 2, 3, \dots$. If for some positive integers $m, n$ we have $T_m = \frac{1}{n}$ and $T_n = \frac{1}{m}$,then $T_{mn}$ equals

Let $a_1, a_2, a_3, \ldots, a_n$ be positive real numbers. Then the minimum value of $\frac{a_1}{a_2}+\frac{a_2}{a_3}+\ldots+\frac{a_n}{a_1}$ is

$A$ person is to count $4500$ currency notes. Let $a_n$ denote the number of notes he counts in the $n^{th}$ minute. If $a_1 = a_2 = \ldots = a_{10} = 150$ and $a_{10}, a_{11}, \ldots$ are in an $A.P.$ with common difference $-2$,then the time taken by him to count all notes is ............... $minutes$.