In an $A.P.$,the sixth term is $a_6 = 2$. If the product $a_1 a_4 a_5$ is the greatest,then the common difference of the $A.P.$ is equal to

If $a_1, a_2, a_3, . . . , a_n, . . .$ are in $A.P.$ such that $a_4 - a_7 + a_{10} = m$,then the sum of the first $13$ terms of this $A.P.$ is .............. $m$.

For $x \geq 0$,the least value of $K$,for which $4^{1+x}+4^{1-x}$,$\frac{K}{2}$,and $16^{x}+16^{-x}$ are three consecutive terms of an $A.P.$ is equal to :

If $\{a_{i}\}_{i=1}^{n}$,where $n$ is an even integer,is an arithmetic progression with common difference $d=1$,and $\sum_{i=1}^{n} a_{i}=192$,$\sum_{i=1}^{n/2} a_{2i}=120$,then $n$ is equal to

Let $\frac{1}{x_1}, \frac{1}{x_2}, \frac{1}{x_3}, \dots, \frac{1}{x_n}$ ($x_i \neq 0$ for $i = 1, 2, \dots, n$) be in $A.P.$ such that $x_1 = 4$ and $x_{21} = 20$. If $n$ is the least positive integer for which $x_n > 50$,then $\sum_{i=1}^n \left( \frac{1}{x_i} \right)$ is equal to:

How many terms of the $A.P.$ $-6, -\frac{11}{2}, -5, \ldots$ are needed to give the sum $-25$?