The common difference of the $A.P.$ $b_{1}, b_{2}, \ldots, b_{m}$ is $2$ more than the common difference of $A.P.$ $a_{1}, a_{2}, \ldots, a_{n}$. If $a_{40} = -159$,$a_{100} = -399$ and $b_{100} = a_{70}$,then $b_{1}$ is equal to:

If ${a_1}, {a_2}, ..., {a_n}$ are positive real numbers whose product is a fixed number $c$,then the minimum value of ${a_1} + {a_2} + ... + {a_{n-1}} + 2{a_n}$ is

In an arithmetic progression,if $S_{40} = 1030$ and $S_{12} = 57$,then $S_{30} - S_{10}$ is equal to:

The sum of the first $p, q,$ and $r$ terms of an $A.P.$ are $a, b,$ and $c,$ respectively. Prove that $\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0$.

If $\sum_{r=1}^n(2r+1)=440$,then $n = \ldots$.

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