If the sum of two extreme numbers of an $A.P.$ with four terms is $8$ and the product of the remaining two middle terms is $15$,then the greatest number of the series will be:

Six numbers are in an $AP$ such that their sum is $3$. The first term is $4$ times the third term. Then, the fifth term is:

If $\frac{a^{n}+b^{n}}{a^{n-1}+b^{n-1}}$ is the $A.M.$ between $a$ and $b,$ then find the value of $n$.

The sum of $n$ terms of an arithmetic progression is $nA + n^2B$,where $A$ and $B$ are constants. The common difference of this progression is.......

Let ${a_1}, {a_2}, \dots, {a_{49}}$ be in $A.P.$ such that $\sum_{k = 0}^{12} {a_{4k + 1}} = 416$ and ${a_9} + {a_{43}} = 66$. If $\sum_{r = 1}^{17} a_r^2 = 140m$,then $m = \dots$

Let $a_1, a_2, a_3, \ldots, a_{11}$ be real numbers satisfying $a_1=15$,$27-2a_2 > 0$,and $a_k = 2a_{k-1} - a_{k-2}$ for $k = 3, 4, \ldots, 11$. If $\frac{a_1^2 + a_2^2 + \ldots + a_{11}^2}{11} = 90$,then the value of $\frac{a_1 + a_2 + \ldots + a_{11}}{11}$ is equal to