If the $n^{th}$ terms of two $A.P.$'s are $3n + 8$ and $7n + 15$,then the ratio of their $12^{th}$ terms will be:

If $1, \log _9(3^{1-x}+2), \log _3(4 \cdot 3^x-1)$ are in $A.P.$,then $x$ equals

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 $a_n, n \geq 1$,be an arithmetic progression with first term $2$ and common difference $4$. Let $M_n$ be the average of the first $n$ terms. Then the sum $\sum_{n=1}^{10} M_n$ 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.

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