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:

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:

The sums of $n$ terms of three $A.P.'s$ whose first term is $1$ and common differences are $1, 2, 3$ are ${S_1}, {S_2}, {S_3}$ respectively. The true relation is

What is the arithmetic mean of the series $a + (a + d) + (a + 2d) + \dots + (a + 2nd)$?

Suppose that all the terms of an arithmetic progression $(A.P.)$ are natural numbers. If the ratio of the sum of the first seven terms to the sum of the first eleven terms is $6:11$ and the seventh term lies between $130$ and $140$,then the common difference of this $A.P.$ is

If the first term of an arithmetic progression is $2$ and the common difference is $4$,then the sum of its first $40$ terms is........