In an $\mathrm{A.P.}$ if $m^{\text {th }}$ term is $n$ and the $n^{\text {th }}$ term is $m,$ where $m \neq n$, find the ${p^{th}}$ term.

Let the sum of the first $n$ terms of a non-constant $A.P., a_1, a_2, a_3, ……$ be $50\,n\, + \,\frac{{n\,(n\, - 7)}}{2}A,$ 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

If the sum of the first $n$ terms of a series be $5{n^2} + 2n$, then its second term is

If $x=\sum \limits_{n=0}^{\infty} a^{n}, y=\sum\limits_{n=0}^{\infty} b^{n}, z=\sum\limits_{n=0}^{\infty} c^{n}$, where $a , b , c$ are in $A.P.$ and $|a| < 1,|b| < 1,|c| < 1$, $abc \neq 0$, then

Let $\frac{1}{{{x_1}}},\frac{1}{{{x_2}}},\frac{1}{{{x_3}}},.....,$  $({x_i} \ne \,0\,for\,\,i\, = 1,2,....,n)$  be in $A.P.$  such that  $x_1 = 4$ and $x_{21} = 20.$ If $n$  is the least positive integer for which $x_n > 50,$  then $\sum\limits_{i = 1}^n {\left( {\frac{1}{{{x_i}}}} \right)} $  is equal to.

In an $A.P.,$ the first term is $2$ and the sum of the first five terms is one-fourth of the next five terms. Show that $20^{th}$ term is $-112$