If the sum of $\mathrm{n}$ terms of an $\mathrm{A.P.}$ is $n P+\frac{1}{2} n(n-1) Q,$ where $\mathrm{P}$ and $\mathrm{Q}$ are constants, find the common difference.

If three positive numbers $a, b$ and $c$ are in $A.P.$ such that $abc\, = 8$, then the minimum possible value of $b$ is

The houses on one side of a road are numbered using consecutive even numbers. The sum of the numbers of all the houses in that row is $170$ . If there are at least $6$ houses in that row and $a$ is the number of the sixth house, then

If twice the $11^{th}$ term of an $A.P.$ is equal to $7$ times of its $21^{st}$ term, then its $25^{th}$ term is equal to

For $p, q \in R$, consider the real valued function $f ( x )=( x - p )^{2}- q , x \in R$ and $q >0$. Let $a _{1}, a _{2}, a _{3}$ and $a _{4}$ be in an arithmetic progression with mean $P$ and positive common difference. If $\left| f \left( a _{ i }\right)\right|=500$ for all $i=1,2,3,4$, then the absolute difference between the roots of $f ( x )=0$ is.

Let ${a_1},{a_2},{a_3}, \ldots $ be terms of $A.P.$  If $\frac{{{a_1} + {a_2} + \ldots + {a_p}}}{{{a_1} + {a_2} + \ldots + {a_q}}} = \frac{{{p^2}}}{{{q^2}}},p \ne q$ then $\frac{{{a_6}}}{{{a_{21}}}}$ equals