The $r$-th term of an arithmetic progression is $T_r$. Its first term is $a$ and the common difference is $d$. If for some positive integers $m, n, m \neq n,$ we have $T_m = 1/n$ and $T_n = 1/m,$ then $a - d = \dots\dots.$

How many multiples of $5$ are there from $10$ to $95$,including both $10$ and $95$?

Let $S_n$ denote the sum of the first $n$ terms of an $A.P$. If $S_4 = 16$ and $S_6 = -48$,then $S_{10}$ is equal to

Let the terms of an arithmetic progression be $a_1, a_2, a_3, \dots$. If $\frac{a_1 + a_2 + \dots + a_p}{a_1 + a_2 + \dots + a_q} = \frac{p^2}{q^2}$,where $p \neq q$,then $\frac{a_6}{a_{21}} = \dots$

Let the sum of the first $n$ terms of a non-constant $A.P.$,$a_1, a_2, a_3, \dots$ be $S_n = 50n + \frac{n(n - 7)}{2}A$,where $A$ is a constant. If $d$ is the common difference of this $A.P.$,then the ordered pair $(d, a_{50})$ is equal to

Let $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}$ be an $A$.$P$. of four terms such that each term of the $A$.$P$. and its common difference $l$ are integers. If $\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}=48$ and $\alpha_{1}\alpha_{2}\alpha_{3}\alpha_{4}+l^{4}=361$, then the largest term of the $A$.$P$. is equal to