Let $S_n$ be the sum of the first $n$ terms of an arithmetic progression $3, 7, 11, \ldots$. If $40 < \left(\frac{6}{n(n+1)} \sum_{k=1}^{n} S_{k}\right) < 42$,then $n$ equals

The sums of $n$ terms of two arithmetic progressions are in the ratio $5n+4 : 9n+6$. Find the ratio of their $18^{th}$ terms.

If the sum of $n$ terms of an $A$.$P$. is $3n^2 + 5n$ and its $m$th term is $164$, then the value of $m$ is

If $\frac{1}{p + q}, \frac{1}{r + p}, \frac{1}{q + r}$ are in $A.P.$,then

If the $p^{th}$ term of an arithmetic progression is $q$ and the $q^{th}$ term is $p$,then its $n^{th}$ term is:

For three positive integers $p, q, r$,$x^{pq p^2} = y^{qr} = z^{p^2 r}$ and $r = pq + 1$ such that $3, 3 \log_y x, 3 \log_z y, 7 \log_x z$ are in $A$.$P$. with common difference $\frac{1}{2}$. Then $r - p - q$ is equal to