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

In an $A.P.$,if the $p^{\text{th}}$ term is $\frac{1}{q}$ and the $q^{\text{th}}$ term is $\frac{1}{p}$,prove that the sum of the first $pq$ terms is $\frac{1}{2}(pq+1)$,where $p \neq q$.

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 the sum of $n$ terms of an $AP$ is given by $S_{n} = n^{2} + n$,then the common difference of the $AP$ is

The sum of the first four terms of an $A.P.$ is $56$. The sum of the last four terms is $112$. If its first term is $11$,then find the number of terms.

If $n$ is a natural number,then $\left( \frac{n+1}{2} \right)^n \ge n!$ is true when