If $a_1, a_2, \dots, a_n$ are positive real numbers such that their product is a constant $c$,then what is the minimum value of $a_1 + a_2 + \dots + a_{n-1} + 2a_n$?

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

If the ratio of the sum of $n$ terms of two $A.P.s$ is $(7n + 1):(4n + 27)$,then the ratio of their $11^{th}$ terms is:

If the sum of the first $11$ terms of an $A.P.$,$a_{1}, a_{2}, a_{3}, \ldots$ is $0$ $(a_{1} \neq 0)$,then the sum of the $A.P.$,$a_{1}, a_{3}, a_{5}, \ldots, a_{23}$ is $k a_{1}$,where $k$ is equal to

If $a_{1}, a_{2}, a_{3}, \ldots$ and $b_{1}, b_{2}, b_{3}, \ldots$ are $A.P.$ and $a_{1}=2, a_{10}=3, a_{1}b_{1}=1=a_{10}b_{10}$,then $a_{4}b_{4}$ is equal to

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